You CAN Ace Calculus
used to prove convergence | yes |
used to prove divergence | no |
can be inconclusive | yes |
inconclusive if all conditions are not met | |
requires the use of limits at infinity | |
\(a_n\) must be positive for \(n>N\) where \(N>0\) | |
showing \( a_{n+1} \leq a_n \) can be tricky |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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free ideas to save on books |
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Help Keep 17Calculus Free |
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The Alternating Series Test is sometimes called the Leibniz Test or the Leibniz Criterion.
Alternating Series Test |
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\( \displaystyle{\sum_{n=1}^{\infty}{(-1)^n a_n}} \) converges, if both of the following conditions hold |
Condition 1 |
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\( \displaystyle{\lim_{n \to \infty}{a_n} = 0} \) |
Condition 2 |
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\( 0 < a_{n+1} \leq a_n \) for all n greater than some \(N>0\) |
Alternate Form Of The Alternating Series TestIn the above theorem, there is a subtle requirement that may not be obvious at first. Notice that in condition 2, the theorem requires \( 0 < a_{n+1} \leq a_n \). There are two things going on. |
What The Alternating Series Test Says |
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Using Mathematical Logic, we can summarize this theorem with the implication
\(\displaystyle{\lim_{n \to \infty}{a_n} = 0}\) ∧
\(\displaystyle{0 < a_{n+1} \leq a_n ~~~ \to ~~~ \sum_{n=1}^{\infty}{(-1)^n a_n} }\) converges
Important Note: Notice the theorem says nothing about divergence. When reading a theorem, you need to look at what it is NOT saying as well as what it IS saying. Since divergence is not part of this theorem, you cannot use this theorem to prove divergence. So you should never say something like, 'the series diverges by the alternating series test.'
When To Use The Alternating Series Test |
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An Alternating Series is exactly what the name implies, i.e. the signs of consecutive elements are opposite. Sometimes this is written with \( (-1)^n \). Other times it may be written with \( (-1)^{n+1} \). And I've even seen it written \( (-1)^{n-1} \). (See the practice problems for examples of additional ways this is written.) Which one is used does not affect convergence or divergence. It will only affect the result if you have a way of determining what the series converges to. The point is that if one element is positive, the next one is guaranteed to be negative (and vice-versa).
The alternating nature of the series is required in order to use this test. Just because a series has negative and positive numbers does not guarantee that is an alternating series. It helps to write out a few terms (5-10) to make sure it is alternating before using this test.
When NOT To Use The Alternating Series Test |
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So when can't we use this test? There are two things to watch for.
1. If either of the conditions fails, this test will not apply. Said another way, all the conditions must hold to be able to apply the test.
2. This test can be used only to test for convergence. If any of the conditions fails, the series may converge or diverge, we just don't know. We need to use another test to determine if it converges or diverges.
In particular, when the first condition fails, we know what test to try next. When we use this test, we usually test the conditions in the order listed above because, if the first condition fails, we are done. Notice the first condition is the
Divergence Test. If it fails, the series diverges (by the divergence test) and we are done.
How To Use The Alternating Series Test |
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To use this test, first make sure you have an alternating series (alternating signs on consecutive terms). Then test condition 1. If it holds, then go on to condition 2 ( and 3 for the alternate form ). How do you show that \( 0 < a_{n+1} \leq a_n \) for all n? There are several ways.
Technique 1 - Directly
First, directly. In this case, set up the inequality and perform algebraic operations until you get an inequality that always holds. For example, we can show that \( n \leq n+2 \) by subtracting n from both sides to get \( 0 \leq 2 \). This last inequality is always true.
Technique 2 - Prove an inequality is always positive.
If the direct way is not possible, try moving all the terms to one side leaving zero on the other side and then explain how the expression is always positive. For example, if we can get something like \( 0 \leq \displaystyle{ \frac{n+5}{n^3} }\) we can argue that, since n is always positive, both the numerator and denominator are positive, resulting in the right side always being greater than zero. The key to remember here is that n starts at zero or one and is always positive after that.
This technique also works if you can find a value N such that the expression holds for all \( n \geq N \). Similar to the last example, you can use this argument for the inequality \( 0 \leq \displaystyle{ \frac{n-5}{n^3} }\)by saying that for \( n \geq 6 \), the inequality holds.
Technique 3 - Using Slope
The third technique is to use the concept of slope and is best demonstrated with an example. Let's show that \( n \geq \ln(n) \) using slope.
The first thing to remember about slope is, to find the slope, you take the derivative and the derivative is defined only on continuous functions. In our case \(n\) is discrete (\(n\) takes on the discrete values 1, 2, 3, 4, ... but no value in between these numbers.), so we need to find continuous functions that have the same values at the discrete values. We don't care what is going on between the discrete values as long as the function is continuous. So for \(n\), we can use \(f(x)=x\) and for \( \ln(n) \) we can use \( g(x)=\ln(x) \) where \(x\) is a continuous variable in both functions. Now we have continuous functions so that we can take derivatives. I know this is a fine point, but we need to get it right.
Okay, we need to show that \( x \geq \ln(x) \)for all \(x\) greater than some value. Let's call \( f(x) = x \)and \( g(x) = \ln(x) \). If we can show that \( f(x) \geq g(x) \) for some specific value of \(x\) and the slope of \( f(x) \) is greater than the slope of \( g(x) \), then \( f(x) \) will always be greater than \( g(x) \). The graphs will never cross and the inequality \( x \geq \ln(x) \) will hold. You can see this intuitively in the graph on the right.
Let's see if we can show this. First, we know that when \( x = 1 \), \( f(1) = 1 \)and \( g(1) = 0 \). Since \( 1 \geq 0 \), we have established a point (\(x=1\)) where \( f(x) \geq g(x) \). Now we will use slope to establish that the functions stay that way for all \( x > 1 \).
Taking the derivatives, we have
\( f'(x) = 1\) and
\( g'(x) = 1/x \).
We need to show \( f'(x) \geq g'(x) \)
\(\displaystyle{
\begin{array}{rcl}
f'(x) & \geq & g'(x) \\
1 & \geq & 1/x \\
x & \geq & 1
\end{array}
}\)
Now, since, \(x\) is always greater than or equal to \(1\), then the slope of \(f(x)\) is always larger than the slope of \(g(x)\). This says that \(f(x)\) is increasing at a faster rate than \(g(x)\) and therefore will always be larger.
To recap, what we have done here is that we have found a point, \(x=1\) where the inequality holds. Then we have used slope to say that the inequality holds for all values greater than that value.
Before jumping into practice problems, here are a couple of video clips to watch. They are only a few minutes long but they may really help you understand how to use this test.
This video clip shows a great explanation of the alternating series test. The instructor uses the alternate form of the test.
video by Dr Chris Tisdell
This next video clip also has a quick explanation of the alternating series test. Watching this will give you another perspective. One thing he leaves out here is that the terms must satisify the inequality \(b_n > 0\).
video by PatrickJMT
Absolute And Conditional Convergence |
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Absolute and conditional convergence applies to all series whether a series has all positive terms or some positive and some negative terms (but the series is not required to be alternating).
One unique thing about series with positive and negative terms (including alternating series) is the question of absolute or conditional convergence. Once convergence of the series is established, then determining the convergence of the absolute value of the series tells you whether it converges absolutely or conditionally. Formally, here's what it looks like.
Definitions of Absolute and Conditional Convergence | ||
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Given that \(\sum{a_n}\) is a convergent series. | ||
if \(\sum{\left|a_n\right|}\) converges |
if \(\sum{\left|a_n\right|}\) diverges | |
then \(\sum{a_n}\) converges absolutely |
then \(\sum{a_n}\) converges conditionally |
Here is a table that summarizes these ideas a little differently.
\(\sum{a_n}\) | \(\sum{\left|a_n\right|}\) |
conclusion | |
---|---|---|---|
converges | converges |
\(\sum{a_n}\) converges absolutely | |
converges | diverges |
\(\sum{a_n}\) converges conditionally |
Absolute Convergence Theorem |
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If the series \( \sum{\left|a_n\right|} \) converges, |
Alternatively, it is possible to determine the convergence of the absolute value of the series first. Then, if the absolute value of the series converges, you can use the Absolute Convergence Theorem to say that the alternating series also converges and converges absolutely.
Additionally, if you have a series with some negative terms (but not all) and it is not an alternating series, you can use this theorem to determine convergence. Specifically, if the absolute value of the series converges, then the series will converge. Notice that this theorem says nothing about divergence, so you cannot make any assumption about convergence or divergence if this theorem does not hold.
This next video clip has a great discussion on absolute convergence including using some examples. Notice that he does not use the term conditional convergence. Instead, he just says that the series does not converge absolutely.
video by Dr Chris Tisdell
Here is another short video clip explanation of absolute and conditional convergence.
video by PatrickJMT
Freaky Consequence of Conditionally Convergent Infinite Series |
As you probably know by now, when you start working with numbers out to infinity, strange things happen. This is certainly true of infinite series which are conditionally convergent. The strange thing is that, when you rearrange the sum, you can get different values to which the series converges, i.e. the commutative property of numbers does not hold! Whoa!
In fact, not only can you get different values, it is possible to rearrange a conditionally convergent infinite series in order to get any number we want, including zero and infinity! This is called the Riemann Series Theorem.
Want to know more? Check out this wikipedia page.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
Instructions: - - Unless otherwise instructed, for each of the following series:
1. Determine whether it converges or diverges using the alternating series test, if possible.
2. If it converges, determine what it converges to, if possible.
3. Determine if a convergent series converges absolutely or conditionally.
Basic Problems |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }\) converges conditionally by the alternating series test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }\)
Solution |
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In addition to this video solution, we show a detail solution here. In the video, he doesn't show conditional convergence in this section of the video.
Convergence Or Divergence
This is an alternating series. So we will start with the Alternating Series Test to see if we can prove convergence.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{ \frac{1}{n} } = 0 }\)
So condition 1 holds.
Condition 2: \( 0 < a_{n+1} \leq a_n \)
\(\displaystyle{ a_{n+1} = \frac{1}{n+1} }\)
We will set up an inequality, perform valid algebraic operations and see if we can get an inequality that holds for all \(n\).
\(\displaystyle{
\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\frac{1}{n+1} & \leq & \frac{1}{n} \\
n & \leq & n+1 \\
0 & \leq & 1
\end{array}
}\)
The last inequality holds for all \(n > 0 \). So condition 2 holds. Therefore the series converges by the Alternating Series Test.
Conditional Or Absolute Convergence
Now we need to test for absolute or conditional convergence.
We need to look at the series \( \sum{1/n} \). Notice this is a p-series with \(p=1\), which means that it diverges. So the original series converges conditionally.
video by Dr Chris Tisdell
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }\) converges conditionally by the alternating series test. |
close solution |
\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }\) converges conditionally by the Alternating Series Test and p-series. |
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }\)
Solution |
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Convergence Or Divergence
This is obviously an alternating series. Let's apply the Alternating Series Test to see if it converges.
Condition 1: \(\displaystyle{ \lim_{n \to \infty}{1/\sqrt{n}} = 0 }\) So condition 1 holds.
Condition 2: We need to show that \( 0 < a_{n+1} \leq a_n \).
Well, \( a_{n+1}=1/\sqrt{n+1} \) is always positive ( since \(n>0\)), which means that \(0 < a_{n+1}\). So we just need to show \( a_{n+1} \leq a_n \)
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\frac{1}{\sqrt{n+1}} & \leq & \frac{1}{\sqrt{n}} \\
\sqrt{n} & \leq & \sqrt{n+1} \\
n & \leq & n+1 \\
0 & \leq & 1
\end{array}\)
Since the last inequality is always true, we know that condition 2 holds. Since both conditions hold, the series converges.
Note: In the above computation, we wrote \(\displaystyle{ \sqrt{n} \leq \sqrt{n+1} \to n \leq n+1 }\). Before doing that step, we need to make sure we are justified in squaring each side to remove the square root sign. Since \(n\) and \(n+1\) are both greater than one, squaring each side is allowed.
Absolute Or Conditional Convergence
Looking at the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\), this is a p-series with \( p = 1/2 \), since \(\displaystyle{\sqrt{n} = n^{1/2} }\). Since \( p \leq 1 \) the series diverges. This means that the original series converges conditionally.
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }\) converges conditionally by the Alternating Series Test and p-series. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }\) converges by the alternating series test, and it converges conditionally by the limit comparison test with a p-series. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }\)
Solution |
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If we write the sum as \(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n a_n } }\) then,
\(\displaystyle{ a_n = \frac{1}{3n+7} }\)
Convergence Or Divergence
This is an alternating series. So we will start with the Alternating Series Test to see if we can prove convergence.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{ \frac{1}{3n+7} } = 0 }\)
So condition 1 holds.
Condition 2: \( 0 < a_{n+1} \leq a_n \)
\(\displaystyle{ a_{n+1} = \frac{1}{3(n+1)+7} }\)
We will set up an inequality, perform valid algebraic operations and see if we can get an inequality that holds for all \(n\).
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\frac{1}{3(n+1)+7} & \leq & \frac{1}{3n+7} \\
3n+7 & \leq & 3(n+1)+7 \\
3n & \leq & 3(n+1) \\
n & \leq & n+1 \\
0 & \leq & 1
\end{array}\)
The last inequality holds for all \(n > 0 \). So condition 2 holds. Therefore the series converges by the Alternating Series Test.
Conditional Or Absolute Convergence
Note: We are going to use the limit comparison test to determine absolute or conditional convergence in this problem. If you have not studied that test yet, you can skip this part and then come back here later.
We are going to compare this series with the divergent p-series \( \sum{1/n} \). How did we decide on that? We used idea 1 from the limit comparison test page. Let \(t_n=1/n\) and set up the limit and evaluate it.
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{t_n}{a_n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{3n+7}{1} \frac{1}{n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{3n+7}{n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| 3 + \frac{7}{n} \right|} }\) |
\( \left| 3 + 0 \right| = 3 \) |
So, since the limit is finite and positive (i.e. not zero or infinity) and the series \(\displaystyle{ \sum{t_n} }\) diverges, then the series \(\displaystyle{ \sum{a_n} }\) also diverges.
So, the original series converges conditionally.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }\) converges by the alternating series test, and it converges conditionally by the limit comparison test with a p-series. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2} } }\) converges by the Alternating Series Test and it converges absolutely by a comparison test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2} } }\)
Solution |
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Convergence Or Divergence
This is an alternating series, so to prove convergence we need to show that both of these conditions hold.
Condition 1: \(\displaystyle{ \lim_{n\to\infty}{a_n} = 0 }\)
Condition 2: \( a_{n+1} \leq a_n \) for all \( n \geq N \) for some N
where \(\displaystyle{\sum_{n=0}^{\infty}{\frac{(-1)^n}{(1+n)^2}} = \sum_{n=0}^{\infty}{(-1)^n a_n} }\) and \(\displaystyle{ a_n = \frac{1}{(1+n)^2} }\)
Okay, let's look at Condition 1:
\(\displaystyle{ \lim_{n\to\infty}{a_n} }\) |
\(\displaystyle{ \lim_{n\to\infty}{\frac{1}{(1+n)^2}} }\) |
\(\displaystyle{ \lim_{n\to\infty}{\frac{1}{(1+n)^2} \cdot \frac{1/n^2}{1/n^2}} }\) |
\(\displaystyle{ \lim_{n\to\infty}{\frac{1/n^2}{(1/n+1)^2}} }\) |
Use the Infinite Limits Theorem. |
\(\displaystyle{ \frac{0}{1} = 0 }\) |
So, Condition 1 holds. Let's look at Condition 2. We need to show \( a_{n+1} \leq a_n \).
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\displaystyle{ \frac{1}{(2+n)^2} } & \leq & \displaystyle{ \frac{1}{(1+n)^2} } \\
(1+n)^2 & \leq & (2+n)^2 \\
(1+n) & \leq & (2+n) \\
1 & \leq & 2
\end{array}\)
The last line is true for all n. So Condition 2 also holds and, therefore, the series converges by the Alternating Series Test (AST).
Absolute or Conditional Convergence
Using one of the comparison tests ( direct or limit ), we can show that the series \(\sum{1/(1+n)^2}\) converges. Therefore the original series converges absolutely.
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2} } }\) converges by the Alternating Series Test and it converges absolutely by a comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos (\pi n)}{n} } }\) converges conditionally by the Alternating Series Test and p-series. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n} } }\)
Solution |
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Convergence Or Divergence
Here is a trick that can obscure what is going on with the series. Sometimes you will find sines and cosines in the expression. When you do, take some time to write out for the first few terms. If you do, you can usually see a pattern to simplify the expression. Let's try it here.
\(\begin{array}{lllrr}
n & & \cos( \pi n ) & & \\
n=1 & & \cos(\pi) &=& -1 \\
n=2 & & \cos(2\pi) &=& 1 \\
n=3 & & \cos(3\pi) &=& -1 \\
n=4 & & \cos(4\pi) &=& 1
\end{array}\)
Do you see the pattern? This is a very common in all textbooks I have looked at. They use sines and cosines to generate alternating signs. So the key is to determine if you need \( (-1)^n \) or \( (-1)^{n+1} \). (Although the difference is not important as far as convergence or divergence, good teachers will encourage you to write it correctly).
In this case, we have \( \cos( \pi n) = (-1)^n \). So our series is \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n} } }\)
This is obviously an alternating series. Let's apply the Alternating Series Test to see if it converges.
Condition 1: \(\displaystyle{ \lim_{n \to \infty}{1/n} = 0 }\) So condition 1 holds.
Condition 2: We need to show that \( 0 < a_{n+1} \leq a_n \).
Well, \( a_n=1/n \) is always positive (since n>0). So we just need to show \( a_{n+1} \leq a_n \)
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\displaystyle{ \frac{1}{n+1} } & \leq & \displaystyle{ \frac{1}{n} } \\
n & \leq & n + 1 \\
0 & \leq & 1
\end{array}\)
Since the last inequality is always true, we know that condition 2 holds. Since both conditions hold, the series converges.
Absolute Or Conditional Convergence
Looking at the series
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}\), this is a p-series with \( p = 1 \). Since \( p \leq 1 \) the series diverges. This means that the original series converges conditionally.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos (\pi n)}{n} } }\) converges conditionally by the Alternating Series Test and p-series. |
close solution |
\(\displaystyle{\sum_{n=3}^{\infty}{\left[(-1)^n\cos(\pi/n)\right]}}\)
Problem Statement |
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\(\displaystyle{\sum_{n=3}^{\infty}{\left[(-1)^n\cos(\pi/n)\right]}}\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=3}^{\infty}{ \left[ (-1)^n \cos(\pi/n) \right] } }\) diverges by the divergence test. |
Problem Statement |
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\(\displaystyle{\sum_{n=3}^{\infty}{\left[(-1)^n\cos(\pi/n)\right]}}\)
Solution |
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Convergence Or Divergence
This is obviously an alternating series. So we need to look at both conditions of the alternating series test.
Condition 1: \(\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{ \cos(\pi/n) } = }\) \(\cos(0) = 1 \)
Since the limit is not zero, condition 1 does not hold. So we cannot continue with the alternating series test. However, we can use this result in the divergence test to say that the series diverges.
Note: We cannot say that the series diverges by the alternating series test. The alternating series test can be used only for showing convergence. If the alternating series test fails (either condition 1 or 2 does not hold), then we can say nothing about the series and we need to use another test to show convergence or divergence.
Final Answer |
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The series \(\displaystyle{ \sum_{n=3}^{\infty}{ \left[ (-1)^n \cos(\pi/n) \right] } }\) diverges by the divergence test. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3(-1)^{n}}{n^2+1} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3(-1)^{n}}{n^2+1} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{\frac{3(-1)^{n}}{n^2+1}} }\) converges absolutely by the Alternating Series Test and comparison to a convergent p-series. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3(-1)^{n}}{n^2+1} } }\)
Solution |
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Convergence Or Divergence
This is obviously an alternating series. Let's apply the Alternating Series Test to see if it converges.
Condition 1: The limit we need to evaluate is \(\displaystyle{ \lim_{n \to \infty}{ \frac{3}{n^2+1} } = 0 }\).
So, condition 1 holds.
Condition 2: We need to show that \(0 < a_{n+1} \leq a_n\).
Well, \( a_{n+1}=3/[(n+1)^2+1] \) is always positive ( since \(n>0\)). So we just need to show \(a_{n+1} \leq a_n\).
\( a_{n+1} \leq a_n \) |
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\( \displaystyle{ \frac{3}{(n+1)^2+1} } \leq \displaystyle{ \frac{3}{n^2+1} } \) |
\( n^2 + 1 \leq (n+1)^2 + 1 \) |
\( n^2 \leq n^2 + 2n + 1 \) |
\( -1 \leq 2n \) |
\( -1/2 \leq n \) |
Okay, since \(n\) starts at zero and increases, it is always greater than \(-1/2\). So this inequality holds for all \(n\) and condition 2 holds. Therefore, the original alternating series converges.
Absolute Or Conditional Convergence
Looking at the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+1} }}\), we can compare this to the convergent p-series \(\sum{1/n^2}\) using one of the comparison tests and determine that the series converges. [The details of that comparison is shown in a practice problem on the limit comparison test page.]
This means that the original series converges absolutely.
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{\frac{3(-1)^{n}}{n^2+1}} }\) converges absolutely by the Alternating Series Test and comparison to a convergent p-series. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2^n} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2^n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2^n} } }\) converges absolutely by the alternating series test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2^n} } }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n}{2^n} } }\) converges absolutely by the alternating series test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n^{3/4}} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n^{3/4}} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n^{3/4}} } }\) converges. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n^{3/4}} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\cos(n\pi)}{n^{3/4}} } }\) converges. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}n}{n^3+1} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}n}{n^3+1} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1} n}{n^3+1} } }\) converges. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}n}{n^3+1} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1} n}{n^3+1} } }\) converges. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^nn}{5n+6} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^nn}{5n+6} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n}{5n+6} } }\) diverges by the divergence test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^nn}{5n+6} } }\)
Solution |
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Be careful here. The instructor does not say that the divergence test tells us that this series diverges. He implies, by not saying anything, that the alternating series test tells us this. But that is NOT correct. The alternating series test tells us only about convergence.
video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n}{5n+6} } }\) diverges by the divergence test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{\sqrt{n^2+6n-2}} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{\sqrt{n^2+6n-2}} } }\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n 5^{3/n} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n 5^{3/n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n 5^{3/n} } }\) diverges by the divergence test |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n 5^{3/n} } }\)
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In this video he actually says that this series diverges by the alternating series test. NO! That is incorrect. Notice that the alternating series test says nothing about divergence, only convergence. If you think I am being picky, then check with your teacher. Chances are that you could lose points if you use this logic. Normally, PatrickJMT is really good but no one is perfect. We are just trying to watch out for you, so that you get the highest score possible in your calculus class and understand the material well.
video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ (-1)^n 5^{3/n} } }\) diverges by the divergence test |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{(n+2)!} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{(n+2)!} } }\)
Final Answer |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{(n+2)!} } }\) converges by the alternating series test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{(n+2)!} } }\)
Solution |
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When he needs to show decreasing, he is not very rigorous and just shows some examples. Here is how you show that the terms are decreasing.
We need to show that \(b_{n+1} \leq b_{n}\). So we set up the inequality, do some algebra and get a result that is true for some \(n > 0\).
\(b_{n+1} \leq b_{n} \) |
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\( \displaystyle{ \frac{1}{(n+3)!} } \leq \displaystyle{ \frac{1}{(n+2)!} } \) |
\( 1 \leq \displaystyle{ \frac{(n+3)!}{(n+2)!} } \) |
We can write \( (n+3)! = (n+3)[(n+2)!] \) |
\( 1 \leq \displaystyle{ \frac{(n+3)[(n+2)!]}{(n+2)!} } \) |
\( 1 \leq n+3 \) |
\( -2 \leq n \) |
The last inequality holds for all \(n>0\), so \(b_{n+1} \leq b_{n}\) which means the terms are decreasing.
video by PatrickJMT
Final Answer |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{(n+2)!} } }\) converges by the alternating series test. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-10)^{n}}{n!} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-10)^{n}}{n!} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-10)^{n}}{n!} } }\) converges absolutely. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-10)^{n}}{n!} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-10)^{n}}{n!} } }\) converges absolutely. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{\sqrt[4]{n}} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{\sqrt[4]{n}} } }\)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}n}{n+5} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}n}{n+5} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}n}{n+5} } }\) diverges by the divergence test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}n}{n+5} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}n}{n+5} } }\) diverges by the divergence test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}4^{2n}}{5^n} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}4^{2n}}{5^n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}4^{2n}}{5^n} } }\) diverges. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}4^{2n}}{5^n} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n}4^{2n}}{5^n} } }\) diverges. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} } }\)
Final Answer |
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The series \( \displaystyle{\sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} } }\) converges absolutely. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \( \displaystyle{\sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} } }\) converges absolutely. |
close solution |
\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
Problem Statement |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\) converges absolutely. |
Problem Statement |
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\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\) converges absolutely. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n^2+2} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n}{n^2+2} } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
Intermediate Problems |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n n}{(\ln(n))^2} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n n}{(\ln(n))^2} } }\)
Final Answer |
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The series diverges by the Divergence Test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n n}{(\ln(n))^2} } }\)
Solution |
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Since this is an alternating series, we will start with the Alternating Series Test. (If you haven't studied the Alternating Series Test yet, don't worry. You will still understand most of this solution.)
Condition 1: We need to show \(\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }\).
Since \(\displaystyle{ \sum_{n=2}^{\infty}{(-1)^n a_n} = \sum_{n=2}^{\infty}{ \frac{(-1)^n n}{ (\ln( n ))^2 } } }\) we know that
\(\displaystyle{ a_n = \frac{n}{ (\ln( n ))^2 } }\)
and so we can write \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{n}{(\ln( n ))^2}} }\).
Plugging in infinity directly yields \(\infty / \infty\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{n}{(\ln( n ))^2}} } \) |
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Apply L'Hôpital's Rule. |
\(\displaystyle{ \lim_{n \to \infty}{\frac{1}{(2 \ln( n ))(1/n)}}}\) |
\(\displaystyle{ \lim_{n \to \infty}{\frac{n}{2 \ln( n )}}}\) |
Plugging in infinity for n, we get the indeterminate form \(\infty/\infty\). So we use L'Hôpital's Rule again. |
\(\displaystyle{ \lim_{n \to \infty}{\frac{1}{2/n}}}\) |
\(\displaystyle{ \lim_{n \to \infty}{\frac{n}{2}} = \infty}\) |
Notice that this is actually the Divergence Test which tells us that the series diverges.
Note: You may be tempted to say that the series diverges by the Alternating Series Test. However, the Alternating Series Test can be used only for convergence. So, it is not correct to say that we used it for divergence. See the infinite series table to confirm this.
Final Answer |
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The series diverges by the Divergence Test. |
close solution |
\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n\ln(n)}{n} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n\ln(n)}{n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln( n )}{ n } } }\) converges conditionally by the Alternating Series Test and the Integral Test. (We could have used the Direct Comparison Test instead of the Integral Test.) |
Problem Statement |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n\ln(n)}{n} } }\)
Solution |
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Convergence Or Divergence
Since this is an alternating series, we will start with the Alternating Series Test.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{\ln( n )}{n}} }\). Plugging in infinity directly yields \(\infty / \infty\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{ \lim_{n \to \infty}{\frac{\ln( n )}{n}} = \lim_{n \to \infty}{\frac{1/n}{1}} = 0 }\)
So condition 1 holds.
To test condition 2, we need to show \( a_{n+1} \leq a_n \).
\(\displaystyle{ \frac{\ln(n+1)}{n+1} \leq \frac{\ln( n )}{n} }\)
If we cross multiply, we get \( n \ln(n+1) \leq (n+1)\ln( n ) \)
There is no easy direct way to show this. However, we do have another technique.
We know that \( a_n \) and \( a_{n+1} \) are always positive (greater than zero). So the graph of \(\displaystyle{ f(x) = \frac{\ln(x)}{x} }\) is always above the x-axis for positive x.
If we can show that the slope is decreasing for all x above some value, then this shows that \( a_{n+1} \leq a_n \) for all n greater than that value. Let's try it.
\(\displaystyle{ f'(x) = \frac{x (1/x) - \ln(x)(1)}{x^2} = \frac{1-\ln(x)}{x^2} }\)
Notice we used the quotient rule here.
So, for the derivative, the denominator is always positive (since it is squared). This means the sign of the derivative will be determined by the numerator. We can see that as long as \(\ln(x)\) is greater than one, the numerator will be negative. And, since \(\ln(x)\) is increasing as \(x\) increases, we just need to see if we can find a number where this starts to occur.
So we need to solve \( 1-\ln(x)=0 \to \ln(x)=1 \to e^{\ln(x)} = e^1 \to x = e \).
This says that if \(x\) is greater than \(e\), then \(1-\ln(x)\) will always be negative. So, since \(e \approx 2.71 \), we can say that for \( x \geq 3 \), \(f'(x)\) is always negative. Therefore the function is decreasing and so \( a_{n+1} \leq a_n \). Condition 2 holds.
Conditional Or Absolute Convergence
Now we need to see if it converges absolutely or conditionally.
So what about the series \(\displaystyle{\sum_{n=2}^{\infty}{\frac{\ln( n )}{n}} }\) ?
Let's use the integral test on this one. If our function is \(\displaystyle{ f(x) = \frac{\ln(x)}{x} }\), we need to find a value where this function starts to decrease and continues to decrease on to infinity. Based on our work above, we know this occurs at x=e. So let's choose x starting at x=3 (any number greater than or equal to e will work here). This gives us the improper integral
\(\displaystyle{ \int_{3}^{\infty}{\frac{\ln(x)}{x}dx} = \lim_{b \to \infty}{\int_{3}^{b}{\frac{\ln(x)}{x}dx} } }\)
This step is necessary since integration is defined only for finite numbers. If your teacher doesn't enforce this, I'm sorry but they should.
Okay, we're going to break out the integral so that we don't have to carry along the limit notation.
\(\displaystyle{ \int{\frac{\ln(x)}{x}dx} }\)
Let \( u = \ln(x) \to du = (1/x)dx \)
\(\displaystyle{ \int{u~du} = (1/2)u^2 = (1/2)(\ln(x))^2 }\)
Strictly, we should have a constant after integrating. However, we know we are going to back and substitute the limits, so it's okay to leave it off.
\(\displaystyle{ \lim_{b \to \infty}{\int_{3}^{b}{\frac{\ln(x)}{x}dx} } }\) |
\(\displaystyle{ \lim_{b \to \infty}{\left[ (1/2)(\ln(x))^2 \right]_{3}^{b}} }\) |
\(\displaystyle{ (1/2)\lim_{b \to \infty}{\left[ (\ln(b))^2 - (\ln(3))^2 \right]} }\) |
\( (1/2)(\infty - (\ln(3))^2) = \infty \) |
So the series diverges.
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln( n )}{ n } } }\) converges conditionally by the Alternating Series Test and the Integral Test. (We could have used the Direct Comparison Test instead of the Integral Test.) |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }\) converges absolutely by the Alternating Series Test and the Ratio Test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }\)
Solution |
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This is an alternating series. So we will start with the Alternating Series Test to see if we can prove convergence.
Keep in mind here that \((2n)! \neq 2(n!)\). Click here for a refresher of factorials.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{n!}{(2n)!}} }\).
Let's look at the fraction and see if we can reduce it to help determine the limit.
\(\displaystyle{ \frac{n!}{(2n)!} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot . . . \cdot n}{1 \cdot 2 \cdot 3 \cdot 4 \cdot . . . \cdot n \cdot (n+1) \cdot (n+2) \cdot . . . \cdot (2n)} }\)
All the terms in the numerator cancel with part of the terms in the denominator, leaving
\(\displaystyle{ \frac{1}{(n+1) \cdot (n+2) \cdot . . . \cdot (2n)} }\)
So this limit is zero, meaning that the first condition holds.
Condition 2: \( 0 < a_{n+1} \leq a_n \)
\(\displaystyle{ a_{n+1} = \frac{(n+1)!}{(2n+2)!} }\)
Notice that the denominator here is \([2(n+1)]! = (2n+2)!\)
For \( n \geq 1 \), we know that \( a_{n+1} > 0 \). So we just need to show the second half of the condition 2 inequality.
\( a_{n+1} \leq a_n \) |
\(\displaystyle{ \frac{(n+1)!}{(2n+2)!} \leq \frac{n!}{(2n)!} }\) |
\(\displaystyle{ \frac{(n+1)!}{n!} \leq \frac{(2n+2)!}{(2n)!} }\) |
\( n + 1 \leq (2n+1)(2n+2) \) |
\( n + 1 \leq 4n^2 + 4n + 2n +2 \) |
\( 0 \leq 4n^2 +5n + 1 \) |
Looking at the last inequality, all terms are positive, therefore the expression on the right is greater than zero, condition 2 holds. So the series converges by the Alternating Series Test. Now we need to test for absolute or conditional convergence.
Let's use the Ratio Test.
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)!}{(2n+2)!} \frac{(2n)!}{n!} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{n+1}{(2n+1)(2)(n+1)} \right|} }\) |
\(\displaystyle{ \frac{1}{2} \lim_{n \to \infty}{\left| \frac{1}{2n+1}\right|} = 0 }\) |
Since 0 < 1, the series converges.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }\) converges absolutely by the Alternating Series Test and the Ratio Test. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{ \infty}{\frac{\sin(n)}{n^2} } }\) converges absolutely by the direct comparison test and the absolute convergence theorem. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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The series \(\displaystyle{ \sum_{n=0}^{ \infty}{\frac{\sin(n)}{n^2} } }\) converges absolutely by the direct comparison test and the absolute convergence theorem. |
close solution |
\(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right) } }\)
Problem Statement |
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\(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right) } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right) } }\) converges. |
Problem Statement |
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\(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right) } }\)
Solution |
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video by Krista King Math
Final Answer |
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The series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right) } }\) converges. |
close solution |
Advanced Problems |
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\(\displaystyle{ \sum_{n=3}^{\infty}{ \left[\frac{(-1)^n}{n\ln(n)-1}\right] } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=3}^{\infty}{ \left[\frac{(-1)^n}{n\ln(n)-1}\right] } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=3}^{\infty}{ \left[ \frac{(-1)^n}{n \ln( n ) - 1} \right] } }\) converges conditionally by the Alternating Series Test and the Direct Comparison Test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=3}^{\infty}{ \left[\frac{(-1)^n}{n\ln(n)-1}\right] } }\)
Solution |
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Convergence or Divergence
This is an alternating series. So we need to make sure that both of these conditions hold.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{1}{n \ln( n ) - 1}} = \frac{1}{\infty} = 0}\). So condition 1 holds.
Condition 2: \( 0 < a_{n+1} \leq a_n \)
\(\displaystyle{ a_{n+1} = \frac{1}{(n+1)\ln(n+1)-1} }\)
For \( n \geq 3 \), we know that \( a_{n+1} > 0 \) (make sure you know how to show this). So we just need to show the second half of the condition 2 inequality.
\( a_{n+1} \leq a_n \) |
\(\displaystyle{ \frac{1}{(n+1)\ln(n+1) - 1} \leq \frac{1}{n \ln( n ) - 1} }\) |
\( n \ln( n ) - 1 \leq (n+1)\ln(n+1)-1 \) |
\( n \ln( n ) \leq (n+1)\ln(n+1) \) |
\( 0 \leq n \ln(n+1) + \ln(n+1) - n \ln( n ) \) |
\( 0 \leq n [ \ln(n+1) - \ln( n ) ] + \ln(n+1) \) |
\(\displaystyle{ 0 \leq n \ln \left( \frac{n+1}{n} \right) + \ln( n+1 ) }\) |
If we can show that the last inequality holds for all \( n \geq 3 \), then condition 2 will hold. Let's look at each piece.
We know that n > 0. For \(\displaystyle{ \ln \left( \frac{n+1}{n} \right) }\), since \(\displaystyle{ \frac{n+1}{n} > 1 }\), the natural log is also positive.
Finally, \( \ln( n+1 ) \) is also positive.
So the result of adding and multiplying pieces that are all positive is also positive. So condition 2 holds. Therefore, the alternating series converges.
Conditional or Absolute Convergence
We need to determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \left[ \frac{1}{n \ln( n ) - 1} \right] } }\). This is determined to diverge in a practice problem on the Direct Comparison Test page. Since this diverges, the original series converges conditionally.
Final Answer |
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The series \(\displaystyle{ \sum_{n=3}^{\infty}{ \left[ \frac{(-1)^n}{n \ln( n ) - 1} \right] } }\) converges conditionally by the Alternating Series Test and the Direct Comparison Test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }\) converges absolutely by the Alternating Series Test and the Ratio Test. |
Problem Statement |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }\)
Solution |
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Convergence Or Divergence
This is an alternating series. So we will start with the Alternating Series Test to see if we can prove convergence.
Condition 1: \(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{2^n}{e^n-1}} }\). This is indeterminate, \(\infty/\infty\), so use L'Hôpital's Rule.
\(\displaystyle{ \lim_{n \to \infty}{\frac{2^n}{e^n-1}} = \lim_{n \to \infty}{\frac{(\ln2)2^n}{e^n}} = \ln(2)\lim_{n \to \infty}{\left(\frac{2}{e}\right)^n} = 0 }\).
So condition 1 holds.
Condition 2: \( 0 < a_{n+1} \leq a_n \)
\(\displaystyle{ a_{n+1} = \frac{2^{n+1}}{e^{n+1}-1} }\)
For \( n \geq 1 \), we know that \( a_{n+1} > 0 \). So we just need to show the second half of the condition 2 inequality.
\( a_{n+1} \leq a_n \) |
\(\displaystyle{ \frac{2^{n+1}}{e^{n+1}-1} \leq \frac{2^n}{e^n-1} }\) |
\( 2(e^n-1) \leq e^{n+1} - 1 \) |
\( 2e^n \leq e^{n+1} + 1 \) |
\( 0 \leq e^{n+1} - 2e^n +1 \) |
\( 0 \leq e^n(e-2) + 1 \) |
Looking at the last inequality, \(e^n > 0, e-2 > 0\) and we are adding a positive one. Since each piece is positive, the result is also positive. So condition 2 holds. Therefore the series converges by the Alternating Series Test.
Conditional Or Absolute Convergence
Now we need to test for absolute or conditional convergence.
Let's use the Ratio Test.
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{2^{n+1}}{e^{n+1}-1} \frac{e^n - 1}{2^n} \right|} }\) |
\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{2(e^n-1)}{e^{n+1}-1} \right|} }\) |
We still have an indeterminate form (which you should verify). So use L'Hôpital's Rule. |
\(\displaystyle{ 2 \lim_{n \to \infty}{\left| \frac{e^n}{e^{n+1}} \right|} }\) |
\(\displaystyle{ 2 \lim_{n \to \infty}{\left| \frac{1}{e} \right|} = 2/e }\) |
Conclusion: Since \( 2/e < 1\), the series converges.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }\) converges absolutely by the Alternating Series Test and the Ratio Test. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^{n-3} \sqrt{n}}{n+13} } }\)
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^{n-3} \sqrt{n}}{n+13} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^{n-3}\sqrt{n}}{n+13} } }\) converges conditionally by the Alternating Series Test and comparison to a divergent p-series. |
Problem Statement |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^{n-3} \sqrt{n}}{n+13} } }\)
Solution |
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Convergence Or Divergence
In this example, the exponent on the \(-1\) is \(n-3\). This does not affect convergence and is just a fancy way of writing \(\displaystyle{ (-1)^{n-1} }\) or \(\displaystyle{ (-1)^{n+1} }\).
This is obviously an alternating series. Let's apply the Alternating Series Test to see if it converges.
Condition 1: The limit we need to evaluate is \(\displaystyle{ \lim_{n \to \infty}{ \frac{\sqrt{n}}{n+13} } }\). We can use either algebra or L'Hôpital's Rule. We will use algebra.
\(\displaystyle{ \lim_{n \to \infty}{ \frac{\sqrt{n}}{n+13} } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \frac{n^{1/2}}{n+13} \frac{1/n}{1/n} } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \frac{n^{-1/2}}{1+13/n} } = 0/1 = 0 }\) |
So, condition 1 holds.
Condition 2: We need to show that \(0 < a_{n+1} \leq a_n\).
Well, \( a_{n+1}=\sqrt{n+1}/(n+14) \) is always positive ( since \(n>0\)). So we just need to show \(a_{n+1} \leq a_n\). However, this is the tricky part of this problem.
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\displaystyle{ \frac{\sqrt{n+1}}{(n+1)+13} } & \leq & \displaystyle{ \frac{\sqrt{n}}{n+13} }
\end{array}\)
There are a couple of things we can try at this point to prove that this inequality holds.
1. First, we could leave both sides as they are, take derivatives and show that the derivative of the right side is greater than the derivative of the left side. Then, if we can find a value of \(n\) where the right side is greater than the left side, then we know this inequality holds. We haven't tried this to see if it works but it really seems like a lot of work.
2. We could multiply both sides by the reciprocal of the left (right) side and show that the result is less (greater) then or equal to one. We thought about this but, again, we think we would need the derivative (which would be quite complicated) to complete it, so we abandoned this option before fully exploring it.
3. The option that we decided on is shown below. The idea is to use algebra to try to get a simple expression that we will show is greater than some value. We played around with the algebra quite a bit, so the work below does not show all of our dead ends. It looks like we just came up with this, but we didn't, at least not right away. It took some work.
\( a_{n+1} \leq a_n \) |
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\(\displaystyle{ \frac{\sqrt{n+1}}{(n+1)+13} \leq \frac{\sqrt{n}}{n+13} }\) |
\(\displaystyle{ \frac{\sqrt{n+1}}{\sqrt{n}} \leq \frac{n+14}{n+13} }\) |
\(\displaystyle{ \frac{n+1}{n} \leq \frac{(n+14)^2}{(n+13)^2} }\) |
\( (n+13)^2 (n+1) \leq n (n+14)^2 \) |
\( 169 \leq n^2 + n \) |
There is a lot of algebra we left out between the last two steps. But I think you can follow what is going on here. We need to be careful when squaring both sides to make sure that the result is valid. It is, in this case, since both sides are greater than one.
Okay, so what do we do now? Since \(13^2 = 169\), we know that for all \(n>13\), the last inequality holds. So this means that condition 2 holds and, so, the series converges.
Absolute Or Conditional Convergence
Looking at the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sqrt{n}}{n+13} }}\), we can compare this to the divergent p-series \(\sum{1/n^{1/2}}\) using one of the comparison tests and determine that the series diverges. This means that the original series converges conditionally.
Final Answer |
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The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^{n-3}\sqrt{n}}{n+13} } }\) converges conditionally by the Alternating Series Test and comparison to a divergent p-series. |
close solution |