Assume $f(x) = \sum_{n=1}^{\infty} a_nx^n$ converges on $(−R, R).$ Show $F(x) = \sum_{n=1}^{\infty} a_nx^{n+1}/(n+1)$ is defined on $(−R, R)$ and satisfies $F'(x) = f(x)$.
My attempt: I know that if power serieas converge on on $(-R,R)$ for all $x$ than its derivative also converge in $(-R,R)$. Assume that $f(x)$ is the derivative of some $F(x)$ and by the above it is defined in $(-R,R)$. Also the antiderivative of $f_n(x) = F_n(x)$, so $F'(x) = f(x)$
Asserting that $\sum_{n=1}^\infty a_nx^n$ converges on $(-R,R)$ is equivalent to the assertion that $\limsup_n\sqrt[n]{\lvert a_n\rvert}\leqslant\frac1R$. But$$\limsup_n\sqrt[n]{\left\lvert\frac{a_{n-1}}n\right\rvert}=\limsup_n\sqrt[n]{\lvert a_n\rvert}.$$Therefore, $\sum_{n=1}^\infty\frac{a_{n-1}}nx^n\left(=\sum_{n=0}^\infty\frac{a_n}{n+1}x^{n+1}\right)$ also converges on $(-R,R)$ and you know that if you differentiate it, you get the original power series.