Proving that the derivative of the integral of a power series equals the original power series I've been thinking about the following recently:
If we have a power series $f(x) = \sum_{n=0}^\infty a_n(x-c)^n$ and $F(x)=\sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}$ where $F(x)$ is constructed by integrating $f(x)$ term-by-term, it seems obvious that $F'(x) = f(x)$. 
However, I realized that I think it is obvious because since high school, I've thought of integration and differentiation like "inverse" operations. What I should do instead is think of them based on more precise definitions. So how would someone prove that $F'(x)=f(x)$ using perhaps the limit definition of a derivative or the Fundamental Theorem of Calculus? 
I started with $\lim_{x \to a} \frac{F(x)-F(a)}{x-a}$ where $a$ is in the interval of convergence for $F(x)$ (which should be the same as the interval of convergence of $f(x)$ I think. If not, please let me know). However, there doesn't seem to be nice cancellation between the sums.
Thank you.

Perhaps the F.T.C. will be helpful:
$$f(x) = \sum_{n=0}^\infty a_n(x-c)^n\text{ and }F(x)=\sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}\\
\text{Then, we need }\int f(x)dx=\sum_{n=0}^\infty a_n\int(x-c)^ndx=\sum^\infty_{n=0}\dfrac{(x-c)^{n+1}}{n+1}=F(x)$$
Q.E.D.(?)
