Question about the radius of convergence and the ratio test? I'm studying for my analysis final and we were given a bunch of practice questions about the radius of convergence.  I understand how to apply the ratio test to find the radius of convergence, but I'm not really sure how to show the following:
Show that the radius of convergence R of the power series $\sum a_nx^n$ is given by lim$|\frac{a_{n}}{a_n+1}|$ whenever it exists.
We're also supposed to find:
Give an example of a power series where this limit does not exist.
Thanks in advance.
 A: Think about what the Ratio Test says.  Given a power series
$$\sum_{n=0}^\infty a_n x^n$$
the ratio test says that the series converges absolutely if
$$L = \lim_{n\to\infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| |x| < 1$$
(it may or may not converge absolutely if that is $1$, but this won't affect the radius of convergence)
Equivalently, we have that
$$|x| < \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$
And since $R$ is defined to be the radius of the largest disk (interval) in which the series converges absolutely, this tells us that
$$R = \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$
and if this limit exists and is non-zero, this tells us in particular that the radius of convergence $R$ is finite, and in this case, is given by precisely
$$R = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$
Concerning the example, try $$\sum_{n=0}^\infty n!\cdot x^n$$
In this case, $a_n = n!$, and $R = \lim_{n\to\infty} \left|\frac{n!}{(n+1)!}\right| = 0$, but $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ does not exist, but this yields $R = 0$, and the series converges only at $0$.
If instead, you want a series where the limit $\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|$ doesn't exist, take 
$$\sum_{n=0}^\infty \frac{x^n}{n!}$$
i.e., $a_n = \frac{1}{n!}$, and in this case, $$\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n\to\infty} \left|\frac{(n+1)!}{n!}\right| = \lim_{n\to\infty} (n+1)$$
which fails to exist, but does in fact yield $R = \infty$, since $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 0$ is the reciprocal of the radius of convergence.
A: I think you wrote the radius down incorrectly. The ratio test says that the series will converge absolutely if $$\lim_{n\to\infty}{\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|}=|x|\lim_{n\to\infty}{\left|{a_{n+1}\over a_n}\right|}<1$$ which happens when $$|x|<\lim{\left|a_n\over a_{n+1}\right|}$$ so the radius is that. I'm not sure what level of rigour your class is looking for, but hopefully that helps.
For the second part, you could take $a_n=0$, or less trivially $a_n=\sin n$.
