## 17Calculus - Infinite Series Practice Exam B

##### 17Calculus

This page contains a complete infinite series exam with worked out solutions.

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Exam Details

Time

1 hour

Questions

9

Total Points

70

Tools

Calculator

not allowed

Formula Sheet(s)

table of series tests

Other Tools

none

Instructions:

- For each problem, correct answers are worth 10%. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

Note: Questions 1-8 are given below. Here is question 9.

Question 9 [5 points]

You must successfully* use each of these tests at least once in questions 1 - 8.

 integral test ratio test root test divergence test direct comparison test limit comparison test

*successfully means that the test proved either convergence or divergence;
- it does not count if the test was inconclusive;
- you must use all these tests to get any points on this question

Questions 1-8

Determine the convergence or divergence of these series.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test listed above unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

$$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the Integral Test)

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum{ne^{-2n^2}} }$$ using the Integral Test.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

The series converges by the Integral Test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum{ne^{-2n^2}} }$$ using the Integral Test.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

For the integral test, we need determine where to determine the lower limit of integration after which the function will be decreasing and positive positive. To do this, we find the critical points.
$$\displaystyle{ f(x) = \frac{x}{e^{2x^2}} }$$
$$\begin{array}{rcl} f'(x) & = & \displaystyle{ \frac{e^{2x^2}(1) - x e^{2x^2}(4x)}{e^{4x^2}} } \\ & = & \displaystyle{ \frac{1-4x^2}{e^{2x^2}} } \end{array}$$
We used the quotient rule in the first line. The denominator of the result is always positive. So we only need to look at the numerator to determine where $$f'(x) = 0$$.
$$1-4x^2 = 0 \to x = \pm 1/2$$
The largest critical value is $$+1/2$$, so let's choose $$1$$ but first make sure that $$f(x)$$ is positive and decreasing.
$$\displaystyle{ f'(1) = \frac{1-4}{e^2} = \frac{-3}{e^2} \lt 0 }$$ So the function is decreasing since $$f'(x) < 0$$ for $$x > 1$$ and $$f(x)$$ is positive. So we choose $$x = 1$$ as the lower limit of integration.
$$\int_{1}^{\infty}{ xe^{-2x^2} dx } = \lim_{b\to\infty}{ \int_{1}^{b}{ xe^{-2x^2} dx } }$$
This step is critical. The first integral cannot be evaluated since integration requires a finite interval. Since this is an improper integral, you must introduce the limit in order to integrate it.
Okay, for now, we will drop the limits of integration and evaluate the indefinite integral $$\int{ xe^{-2x^2} dx }$$ Using substitution, we have $$\displaystyle{ u = -2x^2 \to du = -4xdx \to \frac{du}{-4} = xdx }$$ our integral is now $$\int{ e^u \frac{du}{-4} = \frac{-1}{4}e^u = \frac{-1}{4}e^{-2x^2} }$$ Now we will evaluate the definite integral $$\displaystyle{ \lim_{b\to\infty} \left[ \frac{-1}{4}e^{-2x^2} \right]_1^b = }$$ $$\displaystyle{ \lim_{b\to\infty}\left[ \frac{-1}{4}e^{-2b^2} - \frac{-1}{4}e^{-2} \right] = }$$ $$\displaystyle{ \frac{1}{4e^2} }$$ which is finite, so the integral converges and therefore the series converges by the integral test.

The series converges by the Integral Test.

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$$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + . . . }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + . . . }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

$$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + . . . = }$$ $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{2n+1}{n} } }$$
$$\displaystyle{ \lim_{n\to\infty}{ \left[ \frac{2n+1}{n} \right] } = }$$ $$\displaystyle{ \lim_{n\to\infty}\left[ 2 + \frac{1}{n} \right] = 2 }$$. So the series diverges by the divergence test.

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$$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

Since the $$n^2$$ term dominates all the other terms for large $$n$$, we will compare this series to $$\sum{1/n}$$, which is a divergent p-series. Both the Direct Comparison Test and the Limit Comparison Test work. First, let's use the Direct Comparison Test.
Let $$t_n = 1/n$$ and $$a_n = (2n+1)/(n^2+n)$$. Since the test series diverges, we need to set up the inequality $$0 \lt t_n \lt a_n$$. The left inequality holds since $$0 \lt t_n$$. So we just need to show that $$t_n \lt a_n$$ holds for some $$N$$ where $$n \gt N$$.
$$\begin{array}{rcl} \displaystyle{\frac{1}{n}} & < & \displaystyle{\frac{2n+1}{n^2+n}} \\ \displaystyle{\frac{n^2+n}{n}} & < & 2n + 1 \\ n+1 & < & 2n + 1 \\ n & < & 2n \\ 1 & < & 2 \end{array}$$
The last inequality is true for all $$n$$. So the series diverges by the Direct Comparison Test with the divergent p-series $$\sum{1/n}$$.

Using the Limit Comparison Test and using the same test series, $$t_n = \sum{1/n}$$
$$\begin{array}{rcl} \displaystyle{\lim_{n\to\infty}{\frac{a_n}{t_n}}} & = & \displaystyle{\lim_{n\to\infty}{\frac{2n+1}{n^2+n} \frac{n}{1}}} \\ & = & \displaystyle{\lim_{n\to\infty}{\frac{2n^2+n}{n^2+n}}} \\ & = & 2 \end{array}$$
Since $$\displaystyle{ \lim_{n\to\infty}{ \frac{a_n}{t_n}} = 2 \gt 0 }$$, the series diverges by the Limit Comparison Test with the divergent p-series $$\sum{1/n}$$.

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

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } = 2 \sum_{n=2}^{\infty}{ \frac{2^{n}}{n^n} } }$$
Try the Root Test.
$$\begin{array}{rcl} \displaystyle{\lim_{n\to\infty}{ \sqrt[n]{ |a_n| } } } & = & \displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{ \frac{2^n}{n^n} } } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{2}{n} } } = 0 \end{array}$$
Since $$\displaystyle{\lim_{n\to\infty}{ \sqrt[n]{ |a_n| } } } < 1$$, the series converges by the Root Test.
Note: The Ratio Test also works.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2}{e^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2}{e^n} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

If we try the divergence test, we get zero. So the divergence test is indeterminate. Let's try the Ratio Test.
$$\begin{array}{rcl} \displaystyle{\lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| }} & = & \displaystyle{\lim_{n\to\infty}{ \left[ \frac{(n+1)^2}{e^{n+1}} \cdot \frac{e^n}{n^2} \right] } } \\ & = & \displaystyle{\lim_{n\to\infty}{ \left[ \frac{(n+1)^2}{n^2} \cdot \frac{e^n}{e^{n+1}} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \left( 1 + \frac{1}{n} \right)^2 \cdot \frac{1}{e} \right] } = \frac{1}{e} } \end{array}$$
Since $$\displaystyle{\lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } = \frac{1}{e} < 1 }$$, the series converges by the Ratio Test.
The Root Test also works as well as the Limit Comparison Test. The Direct Comparison Test might also work.

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

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

Using partial fraction expansion, we have $$\displaystyle{ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} }$$. If we build a table starting with $$n=2$$, we get the partial sum $$\displaystyle{ S_n = \frac{1}{2} - \frac{1}{n+1} }$$. Taking the limit as n approaches infinity, we get $$\displaystyle{ \lim_{n\to\infty}{S_n} = \frac{1}{2} }$$. So the Telescoping Series converges to $$1/2$$.
Note: You could have used other tests to prove convergence but you would have received only half the points since you would not have been able to determine what the series converges to.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

This is an alternating series, so we could use the alternating series to prove convergence or divergence. However, the instructions say that if we have positive and negative terms and the series converges, we need to determine if it converges conditionally or absolutely. So we will start by determing whether the series $$\displaystyle{ \sum{a_n} }$$ converges or diverges.
We will use the Ratio Test since we have a factorial.
$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } } & = & \displaystyle{ \lim_{n\to\infty}{ \left| \frac{11^{n+1}}{(n+1)!} \cdot \frac{n!}{11^n} \right| } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left| \frac{11}{n+1} \right| } = 0 } \end{array}$$
So the series converges by the Ratio Test. Since $$\sum{ |a_n| }$$ converges, so does the series $$\sum{a_n}$$ and it converges absolutely by the absolute convergence theorem.
The Root Test may also work.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test that you studied unless told to use a specific test.
- You may not use the idea of growth rate as your answer when evaluating limits.

Solution

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } = }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \frac{2^n}{e^n} } = }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \left( \frac{2}{e} \right)^n } }$$
This is a geometric series with $$r = 2/e < 1$$.
We know a geometric series is $$\displaystyle{ \sum_{n=0}^{\infty}{r^n} = \frac{1}{1-r} }$$. However, our series starts with one, not zero. So we can rewrite this as $$\displaystyle{ \sum_{n=1}^{\infty}{r^n} = \frac{1}{1-r} - 1 }$$
So this series converges to $$\displaystyle{ \frac{2e^{-2}}{e-2} }$$.
Note: You could have used another test to prove convergence but you would not have been able to determine the value to which it converges. Done correctly, this would have given you 3 out of the 5 points.

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