# Convergence radius of a primitive of a complex power series expansion

For a complex power series expansion $$f(z)=\sum_{n=0}^{\infty} c_n(z-a)^n$$ with convergence radius $r$, we have that for $|z-a|<r$: $$f'(z)=\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$$ (this is a theorem that I found in the book I use).

This means that the convergence radius of $\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$ is at least as big as that of $\sum_{n=0}^{\infty} c_n(z-a)^n$.

My question is: is the opposite also true? So, if $\sum_{n=1}^{\infty} nc_n (z-a)^{n-1}$ has convergence radius $\bar{r}$, can we state that $\sum_{n=0}^{\infty} c_n(z-a)^n$ has a convergence radius that is $\geq \bar{r}$. My guess is yes, because we can express a integration process as a differentiation process in the complex plane, but I'm not sure. And if yes, how would I go around proving this?

Yes it is true. A way to prove it is using the root test to find the radius of convergence and to note that it gives the same result for $(c_n)_n$ and $((n+1)c_{n+1})_n$.
• Thanks. So, this means that the expansions of $f$ and $f'$ always have the exact same convergence radius? – user161518 Mar 8 '15 at 15:53