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GRE Subject Test: Math : Tests for Convergence

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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 #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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GRE Subject Test: Math : Tests for Convergence

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #2 : Ratio Test

Example Question #2822 : Calculus Ii

All GRE Subject Test: Math Resources

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