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Single Variable Calculus |
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Acceleration Vector |
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Cylindrical Coordinates |
Lagrange Multipliers |
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Infinite sequences are an important introduction to series, since infinite series are built upon the concept of sequences.
Sequence |
Series | |
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A sequence is just a list of items separated by commas. |
With a series, we actually add up some (or all) terms of some sequence. |
Sequences can be infinite (have an infinite number of terms) or finite (have a finite number of terms).
Example 1: \( \{ 1, 2, 3, 4, 5 \} \) |
This is a finite sequence since it has a finite number of elements, i.e. after the number \(5\) there are no more numbers in the sequence and it stops. |
Example 2: \( \{ 1, 4, 9, 16, 25, 36, . . . \}\) |
This is an infinite sequence since it has an infinite number of elements. The infinite part is denoted by the three periods at the end of the list. |
We don't always list the elements of a sequence. If there is a logical pattern, we can write it more compact form as an equation. This idea is discussed in the next section on notation.
Notation |
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Sequences may be written in several different ways. These are the ones you will come across the most in calculus.
1. As a list of items; we showed this in the two examples above.
2. \(\displaystyle{\{a_n\}_{n=1}^{n=\infty}}\) or more simply as \(\displaystyle{\{a_n\}_{1}^{\infty}}\); in example 2 above, we would also include what \(a_n\) was, i.e. \(a_n = n^2\).
3. Similar to the last way, we could write the sequence in example 2 above as \(\displaystyle{\{n^2\}_{n=1}^{n=\infty}}\); in this case, \(a_n\) is implied to be \(a_n = n^2\).
4. You also often see a sequence written without specifying \(n\), i.e \(\{a_n\}\) and n is implied as \(n = 1, 2, 3, . . . \infty\)
5. A sequence can also be built from itself, called a recursive sequence. In these types of sequences, the first few terms are given and the following terms are defined based on those first terms. For example, \(a_1 = 1, ~~ a_{n+1} = 2a_n\).
6. Finally, you can describe a sequence with words. This is shown in practice problem 239 below. This is a very unusual way to do it, so you won't see it much.
The whole idea is to emphasize that a sequence is a LIST of elements. Each element stands alone and does not interact at all with any other element.
Here is quick introduction video clip that will get you started. It has a couple of good examples.
video by PatrickJMT
Arithmetic Sequence |
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One special type of sequence is an arithmetic sequence, which looks like \( \{ a_n \} \) where \( a_n = c+nk \), \(c\) and \(k\) are constants. An example is \( \{ 3, 12, 21, 30, . . . \} \). Notice that in this example, we start with \(3\) and add \(9\) to get the next element.
Geometric Sequence |
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Another special type of sequence is a geometric sequence. This sequence looks like
\( \{ a_n \} = \{ c, cr, cr^2, cr^3, . . . \}\) where the terms are of the form
\( a_n = cr^n \) for \(n=0, 1, 2 . . . \)
\(c\) is a constant
\(r\) is a term that does not contain \(n\).
An example is \( \{ 1, 3, 9, 27, 81, . . . \} \) where we start with \(1\) and multiply by \(3\) to get the next element. Compare this to the arithmetic sequence where we add to get the next element.
Sequence Convergence and Divergence |
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Convergence of a sequence is defined as the limit of the terms approaching a specific real number. In equation terms, it looks like this. For a sequence \(\{a_n\}\) where \( n=\{1, 2, 3, 4, . . . \}\), we calculate \(\displaystyle{ \lim_{n \to \infty}{a_n} = L }\)
Convergence: If \(L\) is a real number, then we say that the sequence \(\{a_n\}\) converges to \(L\).
Divergence: If \(L = \pm \infty\) or the limit cannot be determined, i.e. it is undefined, then the sequence is said to diverge.
Note: The index \(n\) can start at zero instead of one. This is what happened with the geometric sequence shown in the previous section. In actuality, the index can start at any number. However, for consistency, we usually start it at either zero or one.
Using L'Hôpital's Rule |
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When evaluating the limit \(\displaystyle{ \lim_{n \to \infty}{a_n} = L }\), we sometimes need to use L'Hôpital's Rule. In the course of using L'Hôpital's Rule, we will need to take some derivatives. However, derivatives are only defined on continuous functions. And since a sequence is a set of discontinuous values, we can't directly use L'Hôpital's Rule. Fortunately, we have a theorem that will help us.
Theorem |
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If we can find a continuous function, call it \(g(x)\) where \(g(n) = a_n\) for all \(n\), then \(\displaystyle{ \lim_{x \to \infty}{g(x)} = \lim_{n \to \infty}{a_n} }\). |
Finding a function \(g(x)\) is not usually that hard, if \(a_n\) is given in equation terms. Just replace \(n\) with \(x\).
I know this seems kind of picky but you need to get used to following the exact requirements of theorem. That is one reason we suggest that you read through and try to understand proofs in calculus. Okay, let's get down to some details and how to work with sequences. These next two videos will help you a lot.
This video is a bit long but well worth taking the time to watch. It contains a lot of detail about sequences and how to work with them and has plenty of good examples.
video by Dr Chris Tisdell
This video is a continuation of the previous video and is not as long but it discusses the core of sequences, boundedness and convergence behaviour. You need to watch the previous video first to get the context before watching this one.
video by Dr Chris Tisdell
Binomial Theorem |
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The binomial theorem is a way to expand a term in the form \((a+b)^n\) into a series. We do not cover this topic extensively at this time but here is a video to get you started with an explanation of the equation and an example.
video by PatrickJMT
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, write the first few terms of these sequences and determine if they converge or diverge. If not specified, start the index at n=1.
Basic Problems |
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\(\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }\)
Problem Statement |
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Determine convergence or divergence of the sequence \(\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }\)
Final Answer |
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Since the limit \(\displaystyle{ \lim_{n \to \infty}{a_n} }\) is finite, the sequence \(\displaystyle{ a_n = \frac{2n}{6n-5} }\) converges. |
Problem Statement |
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Determine convergence or divergence of the sequence \(\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }\)
Solution |
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To determine if the sequence converges, we evaluate the limit \(\displaystyle{\lim_{n \rightarrow \infty}{a_n} }\). If this limit is finite, then the sequence converges, otherwise it diverges.
To calculate the limit, we can use either L'HÃ´pital's Rule
\(\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2n}{6n-5} } = \lim_{n \rightarrow \infty}{ \frac{2}{6} } = 1/3 }\)
Or, since the numerator and denominator are both polynomials, we can multiply the numerator and denominator by \( 1/n \).
\(\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2n}{6n-5} \cdot \frac{1/n}{1/n} } = \lim_{n \rightarrow \infty}{ \frac{2}{6-5/n} } }\)
Since we know that \(\displaystyle{ \lim_{n \rightarrow \infty}{5/n} = 0 }\) then \(\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2}{6-5/n}} = 2/6 = 1/3 }\)
Final Answer |
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Since the limit \(\displaystyle{ \lim_{n \to \infty}{a_n} }\) is finite, the sequence \(\displaystyle{ a_n = \frac{2n}{6n-5} }\) converges. |
close solution |
\(\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }\)
Problem Statement |
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\(\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }\)
Final Answer |
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The sequence can be written |
Problem Statement |
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\(\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }\)
Solution |
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First, we will build a table to see if we can get an idea of convergence/divergence. This will also give us the first few terms of the sequence.
n | \(a_n\) |
1 | \(\displaystyle{ \frac{1+1}{1^2} = \frac{2}{1} = 2 }\) |
2 | \(\displaystyle{ \frac{2+1}{2^2} = \frac{3}{4} }\) |
3 | \(\displaystyle{ \frac{3+1}{3^2} = \frac{4}{9} }\) |
4 | \(\displaystyle{ \frac{4+1}{4^2} = \frac{5}{16} }\) |
Okay, so it looks like the sequence may be converging to zero, but let's calculate the limit to make sure.
The limit we need to calculate is \(\displaystyle{ \lim_{n \to \infty}{\frac{n+1}{n^2}} }\) and we can either use algebra or L'HÃ´pital's Rule. Let's use algebra.
\(\displaystyle{ \lim_{n \to \infty}{\frac{n+1}{n^2}} }\) |
\(\displaystyle{\lim_{n \to \infty}{\frac{n+1}{n^2} \frac{1/n^2}{1/n^2}} }\) |
\(\displaystyle{\lim_{n \to \infty}{\frac{1/n+1/n^2}{1}} = \frac{0+0}{1} = 0 }\) |
Final Answer |
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The sequence can be written |
close solution |
\(\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }\)
Problem Statement |
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\(\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }\)
Final Answer |
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The sequence \(\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }\) converges. |
Problem Statement |
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\(\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }\)
Solution |
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In order to determine convergence or divergence, we need to evaluate the limit \(\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} } }\). If it converges to a real number, then the sequence also converges. Otherwise it diverges.
\(\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} \frac{1/n^2}{1/n^2} } }\) |
\(\displaystyle{ \lim_{n \to \infty}{ \frac{3-1/n^2}{10/n+4} } = \frac{3}{4} }\) |
Since the limit is a real number, the sequence converges.
Notes:
- In this problem, we divided the numerator and denominator by term that matches the highest term. In this case, the highest term was \(n^2\). This technique works for polynomials and it doesn't matter if the highest term is in the numerator or the denominator.
- We could have used L'Hôpital's Rule to determine this limit instead, since direct substitution yields \(\infty / \infty\) which is indeterminate.
Final Answer |
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The sequence \(\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }\) converges. |
close solution |
Find the limit of the sequence \(\{a_n\}\) where \(\displaystyle{ a_n = \frac{2n}{5n-3} }\).
Problem Statement |
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Find the limit of the sequence \(\{a_n\}\) where \(\displaystyle{ a_n = \frac{2n}{5n-3} }\).
Solution |
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video by Krista King Math
close solution |
Find the limit of the sequence \(\{a_n\}\) where \(\displaystyle{ a_n = \frac{n^2-n+7}{2n^3+n^2} }\).
Problem Statement |
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Find the limit of the sequence \(\{a_n\}\) where \(\displaystyle{ a_n = \frac{n^2-n+7}{2n^3+n^2} }\).
Solution |
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video by Krista King Math
close solution |
Find the nth term of the sequence \(\{ 1, 4, 9, 16 \ldots \}\)
Problem Statement |
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Find the nth term of the sequence \(\{ 1, 4, 9, 16 \ldots \}\)
Solution |
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video by Krista King Math
close solution |
Find a formula for the general term \(a_n\) of the sequence \(\displaystyle{ \left\{ \frac{1}{2}, -\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6}, \ldots \right\} }\)
Problem Statement |
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Find a formula for the general term \(a_n\) of the sequence \(\displaystyle{ \left\{ \frac{1}{2}, -\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6}, \ldots \right\} }\)
Solution |
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video by Krista King Math
close solution |
\( \{ n(n-1) \} \)
Problem Statement |
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\( \{ n(n-1) \} \)
Solution |
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video by PatrickJMT
close solution |
\(\displaystyle{ \left\{ 1 + \left( -\frac{1}{2} \right)^n \right\} }\)
Problem Statement |
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\(\displaystyle{ \left\{ 1 + \left( -\frac{1}{2} \right)^n \right\} }\)
Solution |
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video by Krista King Math
close solution |
Compute \(\displaystyle{ \lim_{n\to\infty}{ \frac{\ln(n)}{n} } }\)
Problem Statement |
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Compute \(\displaystyle{ \lim_{n\to\infty}{ \frac{\ln(n)}{n} } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
Compute \(\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{n} } }\) using the result from the previous problem.
Problem Statement |
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Compute \(\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{n} } }\) using the result from the previous problem.
Solution |
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video by Dr Chris Tisdell
close solution |
Intermediate Problems |
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\(\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }\)
Problem Statement |
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\(\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }\)
Final Answer |
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The sequence can be written \(\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n}\right\}_{n=0}^{n=\infty} = \left\{ -1, \frac{1}{3}, \frac{-1}{9}, \frac{1}{27}, \ldots \right\} }\) and the sequence converges to zero. |
Problem Statement |
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\(\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }\)
Solution |
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First, we will build a table to see if we can get an idea of convergence/divergence. This will also give us the first few terms of the sequence.
n | \(a_n\) |
0 | \(\displaystyle{ \frac{(-1)^1}{3^0} = \frac{-1}{1} = -1 }\) |
1 | \(\displaystyle{ \frac{(-1)^2}{3^1} = \frac{1}{3} }\) |
2 | \(\displaystyle{ \frac{(-1)^3}{3^2} = \frac{-1}{9} }\) |
3 | \(\displaystyle{ \frac{(-1)^4}{3^3} = \frac{1}{27} }\) |
To determine convergence or divergence, we need to calculate the limit
\(\displaystyle{ \lim_{n \to \infty}{\frac{(-1)^n}{3^n}} }\)
As \(n\) increases, the denominator increases also, so we know that \(\displaystyle{ \lim_{n \to \infty}{\frac{1}{3^n}} = 0 }\)
Looking at the numerator, the values are either 1 or -1. Since the numerator is never zero, we end up with the limit being zero ( since any number times zero is zero ).
Note: As mentioned above the numerator is either 1 or -1 and, we can see from the first few terms, that the sign alternates between consecutive terms. This is called an alternating sequences, referring to this alternating sign. This will be important later when you study alternating series.
Final Answer |
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The sequence can be written \(\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n}\right\}_{n=0}^{n=\infty} = \left\{ -1, \frac{1}{3}, \frac{-1}{9}, \frac{1}{27}, \ldots \right\} }\) and the sequence converges to zero. |
close solution |
For the sequence \(\displaystyle{\left\{\frac{1}{2n+3}\right\}}\) determine whether it is increasing, decreasing, monotonic and if it is bounded. Explain your reasoning.
Problem Statement |
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For the sequence \(\displaystyle{\left\{\frac{1}{2n+3}\right\}}\) determine whether it is increasing, decreasing, monotonic and if it is bounded. Explain your reasoning.
Solution |
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When she is trying to prove that the series is decreasing, she uses a circular argument and never really proves it. She starts out well by writing \(\displaystyle{ a_{n+1} \leq a_n }\)
as \(\displaystyle{ \frac{1}{2n+5} \leq \frac{1}{2n+3} }\) and then she just says this is true. It is better to show this definitively like this.
\(\begin{array}{rcl}
\displaystyle{ \frac{1}{2n+5} } & \leq & \displaystyle{ \frac{1}{2n+3} } \\
2n+3 & \leq & 2n+5 \\
3 & \leq & 5
\end{array}\)
Since \(3 \leq 5\) is always true then \(\displaystyle{ a_{n+1} \leq a_n }\) is also always true. This is the correct way to show the inequality holds.
Again when she discusses boundedness, she doesn't show that the inequalities hold for all n. To do this, you need to set up the equalities and manipulate them like we did above.
video by Krista King Math
close solution |
\(\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }\)
Problem Statement |
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\(\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }\)
Final Answer |
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The sequence \(\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }\) diverges. |
Problem Statement |
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\(\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }\)
Solution |
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In order to determine convergence or divergence, we need to evaluate the limit
\(\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }\). Direct substitution yields \(\infty / \infty\). Since the numerator is not a polynomial, the best way to evaluate this limit is to use L'Hôpital's Rule.
Now, strictly, we can't use L'Hôpital's Rule directly on limit \(\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }\). L'Hôpital's Rule works only on continuous functions since derivatives are involved. But we can invoke the theorem that says \(\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{x \to \infty}{g(x)} }\) when we have a function \(g(n) = a_n\) for all \(n\).
So, let's define \(\displaystyle{ g(x) = \frac{e^{3x}}{x^2} }\)
\(\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }\) |
\(\displaystyle{ \lim_{x \to \infty}{ \frac{e^{3x}}{x^2} } }\) |
\(\displaystyle{ \lim_{x \to \infty}{ \frac{3e^{3x}}{2x} } }\) |
\(\displaystyle{\lim_{x \to \infty}{ \frac{9e^{3x}}{2} } = \infty }\) |
In the above work, we had to apply L'Hôpital's Rule twice. We didn't show the work but we tried direct substitution after the first application and found that we still had an indeterminate form. This then allowed us to apply L'Hôpital's Rule a second time.
So we found that the limit was \(\infty\) and, since this is not a finite real number, the sequence is said to diverge.
Final Answer |
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The sequence \(\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }\) diverges. |
close solution |
\(\displaystyle{ \left\{ \frac{2^n}{3^{n+1}} \right\} }\)
Problem Statement |
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\(\displaystyle{ \left\{ \frac{2^n}{3^{n+1}} \right\} }\)
Solution |
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video by PatrickJMT
close solution |
Advanced Problems |
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Write down the first few terms of the sequence \(\displaystyle{ \{a_n\} }\) where \(a_n\) is the nth digit of \(e\) and discuss the convergence or divergence of the sequence.
Problem Statement |
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Write down the first few terms of the sequence \(\displaystyle{ \{a_n\} }\) where \(a_n\) is the nth digit of \(e\) and discuss the convergence or divergence of the sequence.
Final Answer |
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The sequence \( \{a_n\}=\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}\) where \(a_n\) is the nth digit of the irrational number \(e\). This sequence diverges. |
Problem Statement |
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Write down the first few terms of the sequence \(\displaystyle{ \{a_n\} }\) where \(a_n\) is the nth digit of \(e\) and discuss the convergence or divergence of the sequence.
Solution |
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We know from algebra that the irrational number \(e = 2.718281828459 \ldots \). So the sequence is \(\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}\).
Since \(e\) is irrational, it will never stop or repeat. Therefore the sequence diverges.
Final Answer |
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The sequence \( \{a_n\}=\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}\) where \(a_n\) is the nth digit of the irrational number \(e\). This sequence diverges. |
close solution |