I proved:
If $\sum_{n=0}^\infty a_n x^n$ converges for all $x \in (-R,R)$ then the differentiated series $\sum_{n=1}^\infty na_n x^{n-1}$ converges for all $x \in (-R,R)$.
Please could somebody tell me if my proof is correct?
Let $x \in (-R,R)$ and $t$ be any $t$ such that $|x|<t<R$. Then (ratio test):
$$ \lim_{n \to \infty} \left | { a_{n+1} x^{n+1 }\over a_n x^n} \right | = \lim \left | { a_{n+1} {x^{n+1} \over t^{n+1}} t^{n+1}\over a_n {x^n\over t^n} t^n} \right | = \lim \left | { a_{n+1} t^{n+1}\over a_n t^n} \right | \left | {{x^{n+1} \over t^{n+1}} \over {x^n\over t^n}} \right | = K \left | {x \over t} \right |$$
where $K < 1$ because $\sum a_n x^n$ converges and $|{x \over t}| < 1$ because $t > x$. Therefore $ K \left | {x \over t} \right |<1$ and by the ratio test the series $\sum_{n=1}^\infty na_n x^{n-1}$ converges.