Ratio Test

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GRE Quantitative Reasoning › Ratio Test

Questions 1 - 4

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

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

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

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.

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.

We can't conclude when using the ratio test.

The series is convergent.

$\frac{3}{211}$

$\frac{\pi}{e}$

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.

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

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

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

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

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

None of the other answers.

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.)

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.

We cannot conclude about the nature of the series.

It is clearly divergent.

The series is fast convergent.

$\frac{1}{79}$

$\frac{3}{79}$

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,

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