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Calculus 2 : Convergence and Divergence

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Example Questions

Example Question #1 : Convergence And Divergence

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:

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

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

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

None of the other answers.

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

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:

We cannot conclude when we use the ratio test.

$-\infty$

It is convergent.

$\infty$

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 #2 : Ratio Test

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:

The series is fast convergent.

$\frac{1}{79}$

It is clearly divergent.

$\frac{3}{79}$

We cannot conclude about the nature of the series.

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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Example Question #3 : Convergence And Divergence

Using the ratio test,

what can we say about the series.

$\sum_{_n=1}^{\infty} \frac{1}{n^p}$ where is an integer that satisfies:

p> 1

Possible Answers:

$\infty$

We can't conclude when we use the ratio test.

$\frac{4}{5}$

We can't use the ratio test to study this series.

Correct answer:

We can't conclude when we use the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

$L=\lim_{n\rightarrow \infty}\left|\frac{a_n+1}{a_n}\right|$

then if,

1) L<1 the series converges absolutely.

2) L>1 the series diverges.

3) L=1 the series either converges or diverges.

Therefore we need to evaluate,

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

we have,

$a_{n+1}=\frac{1}{(n+1)^p} ,a_{n}=\frac{1}{n^p}$

therefore:

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

We know that

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

and therefore,

$\lim_{n\rightarrow \infty}(\frac{n}{n+1})^p=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #2822 : Calculus Ii

Consider the following series :

$\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Possible Answers:

We can't conclude when using the ratio test.

$\frac{\pi}{e}$

$\frac{3}{211}$

The series is convergent.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

$L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

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

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

$\lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #1 : Ratio Test

Use the ratio test to determine whether the series below is convergent or divergent.

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

Possible Answers:

$\frac{6}{11}$

The series is convegent.

The series is divergent.

$\frac{4}{13}$

Correct answer:

The series is divergent.

Explanation:

To use the ratio test, we will need to compute the ratio

$\frac{a_{n+1}}{a_n}$ . Then $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

We have $a_{n+1}=\frac{(n+1)!}{2^{n+1}}, a_n=\frac{n!}{2^n}$ we have then :

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

Since we can write :

(n+1)!=(n+1)n!

$\frac{a_{n+1}}{a_n}=\frac{n+1}{2}$

Thus $\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty} \frac{2^n(n+1)!}{2^{n+1}n!}=\infty$ and because $\infty >1$ the series must diverge.

Therefore we conclude that the series is divergent.

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Example Question #5 : Convergence And Divergence

Using the ratio test , what can you say about the following series:

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

Possible Answers:

The series is convergent.

The series will converge and diverge when it gets close to $\infty$ .

The series is divergent.

$\frac{sin(1)}{3}$

The series has two limits.

Correct answer:

The series is convergent.

Explanation:

We will use the comparison test to conclude about the convergence of this series. To show that the majorant series is convergent we will have to call upon 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 either converges or diverges.

We note first that,

$|sin(n)|\le 1 \forall n$ where n is positive integer.

We have $\left|\frac{sin(n)}{7^n}\right|\le\frac{1}{7^n}$ . By the comparison test, if we can show that the series $\sum_{n=1}^{\infty}\frac{1}{7^n}$ is convergent, then by the comparison test, the series $\sum_{n=1}^{\infty}\frac{sin(n)}{7^n}$ is also convergent.

We consider now the series : $\sum_{n=1}^{\infty}\frac{1}{7^n}$ . We have:

$\frac{a_{n+1}}{a_n}=\frac{1}{7}$ and since $\frac{1}{7}< 1$ we conclude that the series is convergent by the ratio test.

This shows that our series is convergent.

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Example Question #8 : Ratio Test

Suppose that a series has positive terms.

If $\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ what can we say about the convergency of the series.

Possible Answers:

We can't conclude.

The series is convergent.

We will need to know the explicit formula for $a_{n}$ .

The series is divergent.

We will need to know the first two terms.

Correct answer:

The series is divergent.

Explanation:

We are given that the series has positive terms. We know that

$\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ . This means that $\frac{a_{n+1}}{a_{n}}=3\frac{3^n}{2^n}}$ .

Now we note that $\lim_{n\rightarrow \infty }\frac{3^n}{2^n}=\infty$ .

Then $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

Since $\infty>1$ the series will diverge.

The ratio test lets us conclude that the series is divergent.

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Example Question #8 : Ratio Test

We will consider the following series :

$\sum_{n=2}^{\infty}\frac{1}{ln(n)^\alpha}$ .

What can you say about the nature of this series using the ratio test? Assume that $\alpha > 0$ .

Possible Answers:

The series is convergent.

We need to know the exact value of $\alpha$ .

The series converges to $\frac{1}{\alpha}$ .

The nature of the series depends on $\alpha$ .

We can't conclude about the nature of the series.

Correct answer:

We can't conclude about the nature of the series.

Explanation:

Note that for $n\ge 2$ and $\alpha>0$ the series is always positive.

To be able to use the ratio test, we will have to compute the ratio:

$\frac{a_{n+1}}{a_{n}}$ . Then find $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ . If L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

We have $a_{n+1}=\frac{1}{ln(n+1)^\alpha}$ hence :

$\frac{a_{n+1}}{a_{n}}=\frac{ln(n)^\alpha}{ln(n+1)^\alpha}=\left(\frac{ln(n)}{ln(n+1)}\right)^\alpha$

Therefore :

$\lim_{n\rightarrow \infty } \frac{a_{n+1}}{a_{n}}=1$ since $\lim_{n\rightarrow \infty }(\frac{ln(n)}{ln(n+1)})^\alpha=1$

By the ratio test we cannot conclude about the nature of the series.

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Example Question #1 : Ratio Test

We consider the following series:

$\sum_ {n=1}^{\infty}a_{n}$ where $a_{n}=\frac{1}{n\sqrt[n]{3}}$ .

Using the ratio test what can you say about the nature of the series?

Is it convergent or divergent?

Possible Answers:

We can't conclude using the ratio test.

$\frac{1}{2}$

$\frac{1}{3}$

The series is convergent.

The series is divergent.

Correct answer:

We can't conclude using the ratio test.

Explanation:

We will use the ratio test noting first that the series is positive.

We will compute the ratio:

$\frac{a_{n+1}}{a_n}$ . Note that: $a_{n+1}=\frac{1}{(n+1)\sqrt[n+1]{3}}$

Hence : $\frac{a_{n+1}}{a_n}=\frac{n}{n+1}\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}$

Now we have $\lim_{n\rightarrow \infty} \frac{n}{n+1}=1$ and $\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}=3^{{\frac{1}{n}}-\frac{1}{n+1}}=3^{\frac{1}{n(n+1)}}$

and we have $\lim_{n\rightarrow \infty}3^{\frac{1}{n(n+1)}}=\lim_{n\rightarrow \infty}e^{\frac{1}{n(n+1)}ln(3))}=1$

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

Therefore the ratio test is inconclusive. We will need to use another test .

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Calculus 2 : Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #2 : Ratio Test

Example Question #3 : Convergence And Divergence

Example Question #2822 : Calculus Ii

Example Question #1 : Ratio Test

Example Question #5 : Convergence And Divergence

Example Question #8 : Ratio Test

Example Question #8 : Ratio Test

Example Question #1 : Ratio Test

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