Using the ratio test to determine whether or not a series converges

What is the ratio test, and what does it show?

The ratio test for convergence lets us determine the convergence or divergence of a series $a_n$ using the limit

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

Once we find a value for $L$ , we can say that

the series converges absolutely if $L<1$ .

the series diverges if $L>1$ or if $L$ is infinite.

the test is inconclusive if $L=1$ .

The ratio test is used most often when our series includes a factorial or something raised to the $n$ th power.

Example

Use the ratio test to say whether the series converges or diverges.

$\sum^{\infty}_{n=1}\frac{n^3}{4^n}$

To use the ratio test, we need to solve for the limit

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

and then evaluate the value of $L$ .

$L=\lim_{n\to\infty}\left|\frac{\frac{(n+1)^3}{4^{n+1}}}{\frac{n^3}{4^n}}\right|$

We can drop the absolute value bars since all of our terms will be positive.

$L=\lim_{n\to\infty}\frac{\frac{(n+1)^3}{4^{n+1}}}{\frac{n^3}{4^n}}$

$L=\lim_{n\to\infty}\frac{(n+1)^3}{4^{n+1}}\left(\frac{4^n}{n^3}\right)$

Grouping like bases together, we get

$L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(\frac{4^n}{4^{n+1}}\right)$

$L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(4^{n-(n+1)}\right)$

$L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(4^{-1}\right)$

$L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(\frac14\right)$

$L=\frac14\lim_{n\to\infty}\frac{(n+1)^3}{n^3}$

$L=\frac14\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{n^3}$

$L=\frac14\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{n^3}\left(\frac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)$

$L=\frac14\lim_{n\to\infty}\frac{\frac{n^3}{n^3}+\frac{3n^2}{n^3}+\frac{3n}{n^3}+\frac{1}{n^3}}{\frac{n^3}{n^3}}$

$L=\frac14\lim_{n\to\infty}\frac{1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}}{1}$

$L=\left(\frac14\right)\frac{1+\frac{3}{\infty}+\frac{3}{\infty}+\frac{1}{\infty}}{1}$

$L=\left(\frac14\right)\frac{1+0+0+0}{1}$

$L=\frac14$

Since $L<1$ , we can say that the original series $a_n$ converges absolutely.