Term-by-term integration of power series

Term-by-term integration of power series: Let $f(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$ have radius of convergence $R>0$. Define $F(x)=\sum_{n=0}^\infty \frac{a_n}{n+1}(x-x_0)^{n+1}.$ Then $F$ has the same radius of convergence $R$, and for all $|x-x_0|<R$, $F'(x)=f(x).$ Equivalently, for $x$ with $|x-x_0|<R$, $\int_{x_0}^{x} f(t)\,dt = \sum_{n=0}^\infty \frac{a_n}{n+1}(x-x_0)^{n+1}.$

This result provides an explicit antiderivative of a power series inside its disk/interval of convergence and is used to compute integrals and derive expansions.

Proof sketch: Fix $r<R$. The series for $f$ converges uniformly on $|x-x_0|\le r$, so one may integrate partial sums term-by-term and then pass to the limit using uniform convergence and integration .