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Calculus 2 : Concepts of Convergence and Divergence

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Example Question #21 : Series In Calculus

There are 2 series, $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ , and they are both divergent. Is $\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ convergent, divergent, or inconclusive?

Possible Answers:

Convergent

Inconclusive

Divergent

Correct answer:

Inconclusive

Explanation:

$\sum_{n=1}^{\infty}a_{n}$ is divergent

$\sum_{n=1}^{\infty}b_{n}$ is divergent

However, unlike convergent series in which the sum of convergent series will produce a convergent series, this is not the case for divergent series. Due to the nature of infinite series, adding together 2 divergent series may be divergent, but it may also produce a convergent series. More information is needed.

$\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ is inconclusive

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Example Question #22 : Series In Calculus

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

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

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:

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

$\\=\lim_{n \to \infty} 6*\left(\frac{6}{11}\right)^{n} \\ \\= 6*0 \\ = 0$

The series is inconclusive by the divergence test.

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Example Question #23 : Series In Calculus

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

$\sum_{n=1}^{\infty}\frac{12+6sin(n)}{n}$

Possible Answers:

Convergent

Divergent

Inconclusive

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:

$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{12+6sin(n)}{n}$

$\lim_{n \to \infty} \frac{12}{n} + \lim_{n \to \infty} \frac{6sin(n)}{n} = 0 + 0 = 0$

This series is inconclusive by the divergence test.

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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:

Divergent

Convergent

Inconclusive

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 #31 : Series In Calculus

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:

Convergent

Divergent

Inconclusive

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 #32 : Series In Calculus

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

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

Possible Answers:

Divergent

Absolutely Convergent

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 #34 : Series In Calculus

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

Absolutely Convergent

Divergent

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 #35 : Series In Calculus

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

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

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.

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

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

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 #36 : 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:

Conditionally Convergent

Absolutely Convergent

Divergent

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 : Limits Of Sequences

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

Divergent

Inconclusive

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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Calculus 2 : Concepts of Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Series In Calculus

Example Question #22 : Series In Calculus

Example Question #23 : Series In Calculus

Example Question #14 : Concepts Of Convergence And Divergence

Example Question #31 : Series In Calculus

Example Question #32 : Series In Calculus

Example Question #34 : Series In Calculus

Example Question #35 : Series In Calculus

Example Question #36 : Series In Calculus

Example Question #1 : Limits Of Sequences

All Calculus 2 Resources

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