TITLE: $\sum_{n=1}^{\infty}f(z^n)$ converges uniformly with $f$ holomorphic
QUESTION [6 upvotes]: Let $f$ be an holomorphic function on the unit ball with $f(0)=0$.
  Prove that $\sum_{n=1}^{\infty}f(z^n)$ is uniformly locally
  convergent in the unit ball.

My attemp:
It is suffice to prove that $\sum_{n=1}^{\infty}f(z^n)$ converges uniformly  in any $\overline{B_0(r)}$ with $r<1$.
$f'$ is also an holomorphic function, defined on a compact set and therfore bounded. Let's say $|f'|\leq M$. Then, for every $z\in \overline{B_0(r)}$, $|f(z^n)-f(0)|\leq M|z^n-0|$, $|f(z^n)|\leq M|z^n| = M|z|^n \leq Mr^n$.
The series $\sum_{n=1}^{\infty}r^n$ converges and therefore by the M test our series converges uniformly. Is it correct?
Thanks.

REPLY [3 votes]: Let's do brute force!
If $f(z)=\sum_{n\geq1}a_nz^n$, then as a formal series $$\sum_{k\geq1}f(z^k)=\sum_{n\geq1}\left(\sum_{d\mid n}a_d\right)z^n.$$ If $f$ converges in the unit disc, we know from the Cauchy-Hadamard formula that $\limsup_{n\to\infty}|a_n|^{1/n}\leq1$, and we have to show that then $$\limsup\left|\sum_{d|n}a_d\right|^{1/n}\leq1.$$ One should be able to prove that

If $(x_n)_{n\geq1}$ is a sequence of non-negative real numbers such that $\limsup_{n\to\infty}x_n^{1/n}\leq1$, then $\limsup_{n\to\infty}(x_1+\cdots+x_n)^{1/n}\leq 1$.

and using this our result follows.