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 $r$:

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

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

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

Then we have

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

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

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

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

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

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

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

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