Show that if the power series $\sum_{n=0}^\infty a_n x^n$ has radius of convergence $R$ and if $\lim_{n \to \infty} |a_{n+1}/a_n|$ exists, then the value of this limit is $R$
I think there might be a typo in the problem: I think the conclusion should be that $1/R$ is the limit of the sequence. If so, I think I have a proof, but I'm hoping someone could confirm that that is indeed a typo and that my solution is right.
To prove the theorem, I will use the fact that $R$ is the unique number for which the following hold: (1) $|x| < R$ implies the series converges absolutely; (2) $|x| > R$ implies the series diverges. So, suppose that $|x| < \lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|$ or $|x| \cdot \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|<1$. Hence
$$\lim_{n \to \infty} |\frac{a_{n+1}x^{n+1}}{a_nx^n}| = |x| \cdot \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1$$
so by the ratio test for series, it follows that series $\sum_{n=0}^\infty a_nx^n$ converges absolutely. A similar argument shows that if $|x | > \lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}|$, then the series diverges. By uniqueness of the radius of convergence, it follows that $\lim_{n \to \infty} |\frac{a_{n}}{a_{n+1}}| = R$ or $\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| = \frac{1}{R}$.
I just realized that I didn't considered the case when the limit is $0$...In any event, does this seem like the right approach?