# Prove radius of convergence for general power series.

## Question:

Show that the radius of convergence of $\sum_{k=0}^\infty c_kx^k$ is $\lim_{n\to\infty} \frac{c_n}{c_{n+1}}$ if this limit exists

## My attempt:

The only thought that occurred to me was to attempt the ratio test:

$\lim_{k\to\infty} |\frac{c_{k+1}x^{k+1}}{c_kx^k}| = \lim_{k\to\infty} |\frac{c_{k+1}\not x^k x}{c_k\not x^k}| = x\lim_{k\to\infty}\frac{c_{k+1}}{c_k}$

But I am unsure of this being the correct method, or how to connect what I am doing with the conclusion I'm seeking.

• Please add the definition of $s_n$ to your question. It is probably the sum of the first $n$ terms of the sequence $c_k$? – Justpassingby Dec 8 '15 at 20:37
• Sorry. I didn't notice the discrepancy. I am looking at the question right now; it's not explicitly defined in any of the questions. However, in previous proofs, $s_n$ represents the sum of the first n terms, so I think you are right. Would it be safe to assume that? – lollercide Dec 8 '15 at 20:43
• I would assume it represents the sum of the first $n$ terms without taking $x$ on board. The existence of the limit should not depend on $x.$ – Justpassingby Dec 8 '15 at 20:45

Lecornu, french mathematician, have proved that, if the ratio $c_{n+1}/c_n$ has a limit L then L is a singular point of $f(x)=\sum_{n=0}^\infty c_n x^n$. But this is not necessary the unique singular point on the circle of convergence.
(counterexample: $f(x) = \frac{x}{1-x}+\ln(1+x) =\sum_{m=1}^\infty \left(1+\frac{(-1)^m}{m}\right)x^m$ have two singular points, 1 and -1, but $c_{n+1}/c_n$ converges to 1.)