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Calculus 2 : Series in Calculus

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Example Questions

Example Question #171 : Series In Calculus

Determine whether the series converges, absolutely, conditionally or in an interval.

$\sum_{n=0}^{\infty}\sqrt{n+1}-\sqrt{n}$

Possible Answers:

Converges in an interval

Does not converge at all

Converges conditionally

Converges absolutely

Correct answer:

Converges absolutely

Explanation:

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Example Question #172 : Series In Calculus

Determine whether the series converges

$\sum_{n=0}^{\infty}\frac{1+\cos n}{3^n}$

Possible Answers:

Converges conditionally

Does not converge at all

Converges absolutely

Converges in an interval

Correct answer:

Converges absolutely

Explanation:

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Example Question #2956 : Calculus Ii

Test for convergence

$\sum_{n=1}^{\infty}\frac{1}{n^3}$

Possible Answers:

Converges conditionally

Cannot be determined

Converges in an interval

Converges absolutely

Diverges

Correct answer:

Converges absolutely

Explanation:

Step 1: Recall the convergence rule of the power series:

According to the convergence rule of the power series....

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges as long as p>1

Step 2: Compare the exponent:

Since p=3 , it is greater than . Hence the series converges.

Step 3: Conclusion of the convergence rule

Now, notice that the series isn't an alternating series, so it doesn't matter whether we check for absolute or conditional convergence.

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Example Question #173 : Series In Calculus

Test for convergence

$\sum_{n=0}^{\infty} \frac{n^2}{1+n^3}$

Possible Answers:

Converges absolutely

Converges in an interval

Can't be determined

Diverges

Converges Conditionally

Correct answer:

Converges absolutely

Explanation:

Step 1: Try and look for another function that is similar to the original function:

$\frac{n^2}{1+n^3}$ looks like $\frac{n^2}{n^3}=\frac{1}{n}$

Step 2: We will now Use the Limit Comparison test

$\lim_{n\to\infty}\frac{1}{n}\times\frac{(1+n^3)}{n^2}=\lim_{n\to\infty}\frac{1+n^3}{n^3}=1$

Since the limit calculated, is not equal to 0, the given series converges by limit comparison test

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Example Question #22 : Comparing Series

Does the following series converge or diverge?

$\sum_{n=1}^{\infty }\frac{n^2+3}{n^4+6}$

Possible Answers:

Diverge

Conditionally converge

Absolutely converge

The series either absolutely converges, conditionally converges, or diverges.

Correct answer:

Absolutely converge

Explanation:

The best way to answer this question would be by comparing the series to another series, $b_{n}$ , that greatly resembles the behavior of the original series, $a_{n}$ . The behavior is determined by the terms of the numerator and the denominator that approach infinity at the quickest rate. In this case:

$b_{n}=\sum_{n=1}^{\infty }\frac{n^2}{n^4}=\sum_{n=1}^{\infty}\frac{1}{n^2}$

When this series is simplifies, it simplifies to a series that converges because of the p-test where p=2>1 .

With two series and the confirmed convergence of one of those series, the limit comparison test can be applied to test for the convergence or divergence of the original series. The limit comparison test states that two series will converge or diverge together if:

$\lim_{n \to \infty}\frac{a_{n}}{b_{n}}=L>0$

Specifically:

$\lim_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim_{n \to \infty}\frac{n^2+3}{n^4+6}\cdot \frac{n^4}{n^2}=\lim_{n \to \infty}\frac{n^6+3n^4}{n^6+6n^2}$

This limit equals one because of the fact that:

$\lim_{x \to\infty }\frac{ax^2+bx}{cx^2}=\frac{a}{c}$ if the coefficients come from the same power.

Because the limit is larger than zero, $a_{n}$ and $b_{n}$ will converge or diverge together. Since it was already established that $b_{n}$ converges, the original seies, $a_{n}$ , converges by the limit comparison test.

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Example Question #174 : Series In Calculus

Determine if the following series converges or diverges:

$\sum_{n=1}^{\infty}\frac{1}{n^23^n}$

Possible Answers:

Series converges

Series diverges

Correct answer:

Series converges

Explanation:

$0\leq\frac{1}{n^23^n}\leq\frac{1}{3^n}$ for all ; $\sum_{n=1}^{\infty}\frac{1}{3^n}$ is a sum of geometric sequence with base 1/3.

Therefore, said sum converges.

Then, by comparison test, $\sum_{n=1}^{\infty}\frac{1}{n^23^n}$ also converges.

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Example Question #23 : Comparing Series

What can be said about the convergence of the series $\sum_{n=1}^{\infty }\frac{1}{n^2+n}$ ?

Possible Answers:

Diverges

Converges

Inconclusive

Correct answer:

Converges

Explanation:

Since $\frac{1}{n^2+n}<\frac{1}{n^2}$ for all n>0, and $\frac{1}{n^2}$ converges as a P-Series, we may conclude that $\frac{1}{n^2+n}$ must also converge by the comparison test.

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Example Question #176 : Series In Calculus

If $b_{n}<a_{n}$ and $a_{n},b_{n}>0$ , and it may be said that $\sum_{n=1}^{\infty }a_{n}$ converges, what may be said about $\sum_{n=1}^{\infty }b_{n}$ ?

Possible Answers:

$\sum_{n=1}^{\infty }b_{n}$ Converges by the comparison test.

$\sum_{n=1}^{\infty }b_{n}$ Converges by the ratio test.

$\sum_{n=1}^{\infty }b_{n}$ Diverges by the comparison test.

$\sum_{n=1}^{\infty }b_{n}$ Converges by the test for convergence of geometric series.

$\sum_{n=1}^{\infty }b_{n}$ Diverges by the ratio test.

Correct answer:

$\sum_{n=1}^{\infty }b_{n}$ Converges by the comparison test.

Explanation:

Given two series,

$b_{n}<a_{n}$ and $a_{n},b_{n}>0$

where

$\sum_{n=1}^{\infty }a_{n}$ converges,

the Comparison Test states that the second series

$\sum_{n=1}^{\infty }b_{n}$ must also converge, if and only if, it is smaller than the first.

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Example Question #1 : Types Of Series

Evaluate:

$\sum_{n=0}^{\infty } \frac{3^{n}}{\pi ^{n+1}}$

Possible Answers:

$\pi + \frac{1}{3}$

$3 - \frac{1}{\pi}$

The series diverges.

$\pi - \frac{1}{3}$

$\frac{1}{\pi - 3 \right )}$

Correct answer:

$\frac{1}{\pi - 3 \right )}$

Explanation:

This can be rewritten as

$\sum_{n=0}^{\infty } \frac{3^{n}}{\pi ^{n+1}}$

$= \sum_{n=0}^{\infty }\left ( \frac{1}{\pi} \cdot \frac{3^{n}}{\pi ^{n }} \right )$

$= \sum_{n=0}^{\infty }\frac{1}{\pi} \left ( \frac{3 }{\pi} \right ) ^{n }$

$\pi > 3$ , so $\frac{3}{\pi} < 1$ , making this a convergent geometric series with initial term $\frac{1}{\pi} \left ( \frac{3 }{\pi} \right ) ^{0 } = \frac{1}{\pi}$ and common ratio $\frac{3 }{\pi}$ . The sum is therefore

$= \frac{ \frac{1}{\pi}}{1 - \frac{3}{\pi }} = \frac{1}{\pi \left (1 - \frac{3}{\pi } \right )} = \frac{1}{\pi - 3 \right )}$ .

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Example Question #1 : Types Of Series

Evaluate:

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }}$

Possible Answers:

$\frac{e}{ \pi - e}$

The series diverges.

$\frac{\pi}{ \pi + e}$

$\frac{\pi}{ \pi - e}$

$\frac{e}{ \pi + e}$

Correct answer:

$\frac{\pi}{ \pi - e}$

Explanation:

This can be rewritten as

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }} = \sum_{n=0}^{\infty }\left ( \frac{e}{\pi } \right )^{n}$ .

This is a geometric series with initial term $\left (\frac{e}{\pi } \right )^{0} = 1$ and common ratio $\frac{e}{\pi } \right$ . Since $e < \pi$ , $\frac{e}{\pi } < 1$ , and the series converges to:

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }} = \frac{1}{1 - \frac{e}{\pi }} = \frac{\pi}{ \pi - e}$

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #171 : Series In Calculus

Example Question #172 : Series In Calculus

Example Question #2956 : Calculus Ii

Example Question #173 : Series In Calculus

Example Question #22 : Comparing Series

Example Question #174 : Series In Calculus

Example Question #23 : Comparing Series

Example Question #176 : Series In Calculus

Example Question #1 : Types Of Series

Example Question #1 : Types Of Series

All Calculus 2 Resources

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