The Ratio Test

Recall that $ \sum_{n=1}^{\infty} a_n
$ is a geometric series if and only if $ a_n = a r^{n-1}$ for some fixed $ a$ and $ r$. Here we call $ r$ the common ratio. Notice that the ratio of any two successive terms is $ r$:

$\displaystyle \frac{a_{n+1}}{a_n} = \frac{a r^{n}}{a r^{n-1}} = r.
$

Moreover, we have $ \sum_{n=1}^{\infty} a r^{n-1}$ converges (to $ \frac{a}{1-r}$) if and only if $ \vert r\vert<1$ (and, of course it diverges if $ \vert r\vert\geq 1$).

Example 6.4.5   For example, $ \sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^{n-1}$ converges to $ \frac{3}{1-\frac{2}{3}}=9$. However, $ \sum_{n=1}^{\infty} 3\left(\frac{3}{2}\right)^{n-1}$ diverges.

Theorem 6.4.6 (Ratio Test)   Consider a sum $ \sum_{n=1}^{\infty} a_n
$. Then
  1. If $ \lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = L < 1$ then $ \sum_{n=1}^{\oo } a_n$ is absolutely convergent.
  2. If $ \lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = L > 1$ then $ \sum_{n=1}^{\oo } a_n$ diverges.
  3. If $ \lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = L = 1$ then we may conclude nothing from this!

Proof. We will only prove 1. Assume that we have $ \lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = L < 1$. Let $ r=\frac{L+1}{2}$, and notice that $ L<r<1$ (since $ 0\leq L<1$, so $ 1\leq L+1<2$, so $ 1/2 \leq r < 1$, and also $ r-L = (L+1)/2 -L = (1-L)/2 > 0$).

Since $ \lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n}\right\vert = L$, there is an $ N$ such that for all $ n>N$ we have

$\displaystyle \left\vert\frac{a_{n+1}}{a_n}\right\vert < r,$    so $\displaystyle \quad \vert a_{n+1}\vert < \vert a_n\vert \cdot r .
$

Then we have

$\displaystyle \sum_{n=N+1}^{\infty} \vert a_n\vert < \vert a_{N+1}\vert \cdot \sum_{n=0}^{\oo } r^n.
$

Here the common ratio for the second one is $ r<1$, hence thus the right-hand series converges, so the left-hand series converges. $ \qedsymbol$

Example 6.4.7   Consider $ \displaystyle \sum_{n=1}^{\oo } \frac{(-10)^n}{n!}$. The ratio of successive terms is

$\displaystyle \left\vert \frac{\displaystyle \frac{(-10)^{n+1}}{(n+1)!}}{\displ...
...t
= \frac{10^{n+1}}{(n+1)n!} \cdot \frac{n!}{10^n} = \frac{10}{n+1} \to 0 < 1.
$

Thus this series converges absolutely. Note, the minus sign is missing above since in the ratio test we take the limit of the absolute values.

Example 6.4.8   Consider $ \displaystyle \sum_{n=1}^{\infty} \frac{n^n}{3^{1+3n}}$. We have

$\displaystyle \left\vert\displaystyle \frac{\displaystyle \frac{(n+1)^{n+1}}{3\...
...ac{27^n}{n^n}
= \frac{n+1}{27} \cdot \left(\frac{n+1}{n}\right)^n
\to +\infty
$

Thus our series diverges. (Note here that we use that $ \left(\frac{n+1}{n}\right)^n\to e$.)

Example 6.4.9   Let's apply the ratio test to $ \sum_{n=1}^{\oo } \frac{1}{n}$. We have

$\displaystyle \lim_{n\to\infty} \left\vert\frac{\displaystyle \frac{1}{n+1}}{\d...
...rac{1}{n}}\right\vert
= \frac{1}{n+1} \cdot \frac{n}{1} = \frac{n}{n+1} \to 1.
$

This tells us nothing. If this happens... do something else! E.g., in this case, use the integral test.

William Stein 2006-03-15