# Series with Same Radius of Convergence

I'd like to prove the following proposition:

The power series $\sum_{j=0}^{\infty} a_j(x - c)^j$ and the series $\sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ obtained from term by term integration have the same radius of convergence, and the function $F$ defined by $F(x) = \sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ on the common interval of convergence satisfies $F'(x) = \sum_{j=0}^{\infty} a_j(x - c)^j = f(x)$.

Does this occur as a result of $\sum_{j=0}^{\infty} a_j(x - c)^j$ and $\sum_{j=0}^{\infty} \frac{a_j}{j+1}(x - c)^{j+1}$ being real analytic? (Are they real analytic?)

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Cauchy-Hadamard theorem. –  davin Mar 16 '12 at 9:27

Recall that the radius of convergence $R$ of the series $\sum\limits_na_nx^n$ is characterized by the fact that, for every $|x|\lt R$, $a_nx^n\to0$ geometrically fast while, for every $|x|\gt R$, the sequence $(a_nx^n)_n$ is unbounded.
Introduce $b_n=a_n/(n+1)$ and let $R'$ denote the radius of convergence of the series $\sum\limits_nb_nx^n$.
For every $|x|\lt R$, $a_nx^n\to0$ hence $b_nx^n\to0$. This shows that $R'\geqslant R$. For every $|x|\gt R$, $|a_nx^n|\geqslant1$ infinitely often hence $|b_nx^n|\geqslant1/(n+1)$ infinitely often. In particular, the sequence $(b_nx^n)_n$ does not converge to $0$ geometrically fast, hence $R'\leqslant R$.
Hence $R'=R$.