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Calculus 2 : Series in Calculus

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #14 : Concepts Of Convergence And Divergence

Use the limit test (divergence test) to find if the series is convergent, divergent, or inconclusive.

$\sum_{n=1}^{\infty}\frac{3n^2+4n+6}{11n^2+7n+4}$

Possible Answers:

Inconclusive

Divergent

Convergent

Correct answer:

Divergent

Explanation:

Divergence Test and Limit Test are the same tests with different names.

If $\lim_{n \to \infty} a_{n} \neq 0$ then $a_{n}$ diverges.

However, if $\lim_{n \to \infty} a_{n} = 0$ the test is inconclusive.

Solution:

$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{3n^2+4n+6}{11n^2+7n+4}$

$\lim_{n \to \infty} \frac{3n^2}{11n^2} = \lim_{n \to \infty} \frac{3}{11} = \frac{3}{11}$

$\lim_{n \to \infty} a_{n} = \frac{3}{11} \neq 0$

The series is divergent by the divergence test.

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Example Question #15 : Concepts Of Convergence And Divergence

Use the limit test (divergence test) to find if the series is convergent, divergent, or inconclusive.

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

Possible Answers:

Inconclusive

Convergent

Divergent

Correct answer:

Inconclusive

Explanation:

Divergence Test and Limit Test are the same tests with different names.

If $\lim_{n \to \infty} a_{n} \neq 0$ then $a_{n}$ diverges.

However, if $\lim_{n \to \infty} a_{n} = 0$ the test is inconclusive.

Solution:

$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{ln(n)}{n^3}=0$

Although ln(n) also tends to infinity, n grows to infinity more quickly than ln(n), and so the limit will go to 0.

The series is inconclusive by the divergence test.

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Example Question #16 : Concepts Of Convergence And Divergence

Does the following series converge absolutely, conditionally, or is it divergent?

$\sum_{n=1}^{\infty}(\frac{-9}{20})^n$

Possible Answers:

Absolutely Convergent

Divergent

Conditionally Convergent

Correct answer:

Absolutely Convergent

Explanation:

A series is absolutely convergent if $\sum_{n=1}^{\infty}|a_{n}|$ is convergent.

If $\sum_{n=1}^{\infty}|a_{n}|$ is not convergent, but $\sum_{n=1}^{\infty}a_{n}$ is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

$\sum_{n=1}^{\infty}|a_{n}|= \sum_{n=1}^{\infty}\left(\frac{9}{20}\right)^n$ is a geometric series, with r < 1 .

So this series is absolutely convergent by the geometric test.

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Example Question #11 : Concepts Of Convergence And Divergence

Does the following series converge absolutely, conditionally, or is it divergent?

$\sum_{n=1}^{\infty}\left(\frac{-15}{2}\right)^n$

Possible Answers:

Conditionally Convergent

Divergent

Absolutely Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if $\sum_{n=1}^{\infty}|a_{n}|$ is convergent.

If $\sum_{n=1}^{\infty}|a_{n}|$ is not convergent, but $\sum_{n=1}^{\infty}a_{n}$ is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

$a_{n}=\left(\frac{-15}{2}\right)^n$

$|a_{n}| = \left(\frac{15}{2}\right)^n$

$\sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty}\frac{15}{2}$

This is a geometric series with r > 1, and |r| > 1, so this series is divergent by the geometric test.

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Example Question #18 : Concepts Of Convergence And Divergence

Does the following series converge absolutely, conditionally, or is it divergent?

$\sum_{n=1}^{\infty}\frac{(-7)^n}{n^3}$

Possible Answers:

Divergent

Absolutely Convergent

Conditionally Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if $\sum_{n=1}^{\infty}|a_{n}|$ is convergent.

If $\sum_{n=1}^{\infty}|a_{n}|$ is not convergent, but $\sum_{n=1}^{\infty}a_{n}$ is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

$\sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty} \frac{7^n}{n^3}$

$\lim_{n \to \infty}|a_{n}|=\lim_{n \to \infty}\frac{7^n}{n^3}=\infty$

This series diverges by the divergence test.

Now test for conditional convergence.

$\lim_{n \to \infty}a_{n} = \lim_{n \to \infty} (-1)^n\frac{7^n}{n^3}$

This limit bounces between positive and negative infinity, and so the limit does not exist. By the divergence test, this series diverges.

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Example Question #34 : Introduction To Series In Calculus

Does the following series converge absolutely, conditionally, or is it divergent?

$\sum_{n=1}^{\infty}\frac{-e^n}{n^4}$

Possible Answers:

Absolutely Convergent

Divergent

Conditionally Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if $\sum_{n=1}^{\infty}|a_{n}|$ is convergent.

If $\sum_{n=1}^{\infty}|a_{n}|$ is not convergent, but $\sum_{n=1}^{\infty}a_{n}$ is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

$\lim_{n \to \infty} |a_{n}| = \lim_{n \to \infty}\frac{e^n}{n^4} = \infty$

The absolute series diverges by the divergence test.

Now test for conditional convergence.

$a_{n} = (-1)^n\frac{e^n}{n^4}$

$\lim_{n \to \infty}a_{n}$ alternates between positive and negative infinity, and so it diverges by the divergence test.

This series diverges.

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Example Question #1 : Sequences & Series

There are 2 series $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ and $\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

Is the sum of these 2 infinite series convergent, divergent, or inconclusive?

Possible Answers:

Convergent

Inconclusive

Divergent

Correct answer:

Convergent

Explanation:

A way to find out if the sum of the 2 infinite series is convergent or not is to find out whether the individual infinite series are convergent or not.

Test the first series

$\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ .

This is a geometric series with $r = \frac{3}{7} < 1$ .

By the geometric test, this series is convergent.

Test the second series

$\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

This is a geometric series with $r = \frac{5}{11} < 1$ .

By the geometric test, this series is convergent.

Since both of the series are convergent, $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n + \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ is also convergent.

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Example Question #1 : Ratio Test

Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)

Possible Answers:

None of the other answers.

$\sum_{k=0}^{\infty} k$

$\sum_{k=0}^{\infty} \frac{k^2}{3^k}$

$\sum_{k=0}^{\infty} \frac{(-2)^k}{k!}$

$\sum_{k=0}^{\infty} \frac{1}{k!}$

Correct answer:

$\sum_{k=0}^{\infty} k$

Explanation:

The ratio test fails when $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1$ . Otherwise the series converges absolutely if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | <1$ , and diverges if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1$ .

Testing the series $\sum_{k=0}^{\infty} k$ , we have

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.$

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

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Example Question #1 : Ratio Test

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

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

Possible Answers:

$\infty$

$-\infty$

We cannot conclude when we use the ratio test.

It is convergent.

Correct answer:

We cannot conclude when we use the ratio test.

Explanation:

As required by this question we will have to use the ratio test. $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}$ .

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

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Example Question #1 : Convergence And Divergence

We consider the series : $\sum_{n=1}^{\infty}\frac{1}{72n+89}$ , use the ratio test to determine the type of convergence of the series.

Possible Answers:

$\frac{3}{79}$

It is clearly divergent.

We cannot conclude about the nature of the series.

The series is fast convergent.

$\frac{1}{79}$

Correct answer:

We cannot conclude about the nature of the series.

Explanation:

To be able to use to conclude using the ratio test, we will need to first compute the ratio then use $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge. Computing the ratio we get,

$\frac{a_{n+1}}{a_{n}}$ .

We have then:

$a_{n+1}=\frac{1}{72(n+1)+89}=\frac{1}{72n+161}$

Therefore have :

$\frac{a_{n+1}}{a_{n}}=$ $\frac{72n+89}{72n+161}$

It is clear that $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1$ .

By the ratio test , we can't conclude about the nature of the series.

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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 #14 : Concepts Of Convergence And Divergence

Example Question #15 : Concepts Of Convergence And Divergence

Example Question #16 : Concepts Of Convergence And Divergence

Example Question #11 : Concepts Of Convergence And Divergence

Example Question #18 : Concepts Of Convergence And Divergence

Example Question #34 : Introduction To Series In Calculus

Example Question #1 : Sequences & Series

Example Question #1 : Ratio Test

Example Question #1 : Ratio Test

Example Question #1 : Convergence And Divergence

All Calculus 2 Resources

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