Show that the function $F(x) = \sum_{n=0}^{\infty} \frac{a_{n} x^{n+1}}{n+1} $ is defined on $(-R,R)$, and $F'(x) = f(x)$.
We have not yet formally covered anti differentiation of series in class. I am guessing that the term-by-term differentiability of series will be utilized here. Would I be allowed to write out the terms of the partial sum, divide them by ${n+1}$, then multiply them all by $x$ to form the anti-derivative power series, or is that not allowed?
As per showing $F'(x) = f(x)$, that seems like I can just take the derivative, and there's nothing mysterious about it, correct? (I would use term-by-term differentiation.)
Showing that $F$ is a well-defined function on $(-R,R)$ amounts to proving that the radius of convergence $\rho$ of the series defining $F$ is $R$: $$ \frac{1}{\rho} = \limsup_{n\to\infty}\sqrt[n]{\frac{a_n}{n+1}} = \frac{\limsup_{n\to\infty}\sqrt[n]{a_n}}{\lim_{n\to\infty}\sqrt[n]{n+1}} = \limsup_{n\to\infty}\sqrt[n]{a_n} = \frac{1}{R}, $$ since the radius of convergence of the series $f$ is $R$. Since the radius of convergence of $F$ is $R$, we can differentiate $F$ on its interval of convergence term-by-term to get $F'(x) = \sum_{n\ge 0}a_nx^n = f(x)$, as desired.