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You CAN Ace Calculus

17calculus > infinite series > alternating series

Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

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Alternating Series and Absolute/Conditional Convergence

Alternating Series Test

$$\displaystyle{\sum_{n=1}^{\infty}{(-1)^n a_n}}$$ converges, if both of the following conditions hold

Condition 1

Condition 2

$$\displaystyle{\lim_{n \to \infty}{a_n} = 0}$$

$$0 < a_{n+1} \leq a_n$$
for all n greater than some $$N>0$$

Alternate Form Of The Alternating Series Test

In 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.
First, $$a_{n+1} \leq a_n$$. This condition usually gets the most attention.
Second, $$0 < a_{n+1}$$. This may not be so obvious but it is very important.

To highlight the importance of $$0 < a_{n+1}$$, some mathematicians list three conditions, which may appear in a different order than shown here, by breaking condition 2 above into two conditions. Here is what it looks like.

Condition 1: $$\displaystyle{\lim_{n \to \infty}{a_n} = 0}$$ [ same as above ]
Condition 2: $$a_{n+1} \leq a_n$$
Condition 3: $$0 < a_{n+1}$$

Although we call this an alternate form, it is probably a better form than the one at the top of the page, since it more clearly shows the requirements of the theorem.

Quick Summary

 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

The Alternating Series Test is sometimes called the Leibniz Test or the Leibniz Criterion.

What The Alternating Series Test Says

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

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

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 nth-Term Test. If it fails, the series diverges (by the nth-term test) and we are done.

How To Use The Alternating Series Test

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.

 Dr Chris Tisdell - Alternating series and absolute convergence

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 $$b_n > 0$$.

 PatrickJMT - Alternating Series

Absolute And Conditional Convergence

 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

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

If the series $$\sum{\left|a_n\right|}$$ converges,
then the series $$\sum{a_n}$$ also 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.

 Dr Chris Tisdell - Alternating series and absolute convergence

Here is another short video clip explanation of absolute and conditional convergence.

 PatrickJMT - Absolute Convergence, Conditional Convergence and Divergence
 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.

Search 17Calculus

Practice Problems

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.

 Level A - Basic

Practice A01

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{n} \right] } }$$

solution

Practice A02

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n}{\sqrt{n}} } }$$

solution

Practice A03

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{3n+7} \right] } }$$

solution

Practice A04

$$\displaystyle{\sum_{n=0}^{\infty}{ \frac{(-1)^n}{(1+n)^2}}}$$

solution

Practice A05

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{\cos(n\pi)}{n}}}$$

solution

Practice A06

$$\displaystyle{\sum_{n=3}^{\infty}{\left[(-1)^n\cos(\pi/n)\right]}}$$

solution

Practice A07

$$\displaystyle{\sum_{n=0}^{\infty}{\frac{3(-1)^{n}}{n^2+1}}}$$

solution

Practice A08

$$\displaystyle{\sum_{n=0}^{\infty}{\frac{(-1)^n}{2^n}}}$$

solution

Practice A09

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{\cos(n\pi)}{n^{3/4}}}}$$

solution

Practice A10

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}n}{n^3+1}}}$$

solution

Practice A11

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^nn}{5n+6}}}$$

solution

Practice A12

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{\sqrt{n^2+6n-2}}}}$$

solution

Practice A13

$$\displaystyle{\sum_{n=1}^{\infty}{(-1)^n 5^{3/n}}}$$

solution

Practice A14

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{(n+2)!}}}$$

solution

Practice A15

$$\displaystyle{\sum_{n=0}^{\infty}{\frac{(-10)^{n}}{n!}}}$$

solution

Practice A16

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{\sqrt[4]{n}}}}$$

solution

Practice A17

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n}n}{n+5}}}$$

solution

Practice A18

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^{n}4^{2n}}{5^n}}}$$

solution

Practice A19

$$\displaystyle{\sum_{n=1}^{\infty}{ \frac{\sin(\pi/2+n\pi)}{n^{7/2}} }}$$

solution

Practice A20

$$\displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} }}$$

solution

Practice A21

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^2+2}}}$$

solution

 Level B - Intermediate

Practice B01

$$\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n n}{(\ln(n))^2}}}$$

solution

Practice B02

$$\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n\ln(n)}{n}}}$$

solution

Practice B03

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n!}{(2n)! } } }$$

solution

Practice B04

$$\displaystyle{ \sum_{n=0}^{\infty}{\frac{\sin(n)}{n^2}} }$$

solution

Practice B05

$$\displaystyle{\sum_{k=1}^{\infty}{(-1)^{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}}$$

solution

Practice C01

$$\displaystyle{\sum_{n=3}^{\infty}{\left[\frac{(-1)^n}{n\ln(n)-1}\right]}}$$

solution

Practice C02

$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^n}{e^n - 1 } \right] } }$$

$$\displaystyle{ \sum_{n=0}^{\infty}{\frac{(-1)^{n-3} \sqrt{n}}{n+13}} }$$