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GRE Subject Test: Math : Sequences & Series

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

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Example Questions

Example Question #2 : Convergence And Divergence

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:

$\infty$

We cannot conclude when we use the ratio test.

$-\infty$

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 : 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 #2 : Ratio Test

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:

The series is convergent.

$\frac{3}{211}$

$\frac{\pi}{e}$

We can't conclude when using the ratio test.

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 : Sequences & Series

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #2 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #2 : Ratio Test

All GRE Subject Test: Math Resources

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