Does $\sum_{n=1}^{\infty}z^{n} f(z^{n})$ converge uniformly when $f$ is holomorphic $\newcommand{\compconj}[1]{%
 \overline{#1}%
}$
I have to show that $$\sum_{n=1}^{\infty}g(z^n)$$ converges uniformly on every compact circle $\compconj {K_r(0)}$, $r<1$,  where $g(z)$ is defined as $K_1(0)\to \mathbb{C}, g(z)=zf(z)$ where $f(z)$ has the same domain and is holomorphic and $K_r(0)$ is the open unit circle.
What I did:
By plugging in the definition of $g(z)$ we get $$\sum_{n=1}^{\infty}g(z^n)=\sum_{n=1}^{\infty}z^{n} f(z^{n})$$ and I can interpret this as a power series with coefficients $f(z^n)$. I want to show that its radius of convergence is $1$, then I know that is converges uniformly on every compact unit-circle $\compconj{K_r(0)}$ with $r<1$(That is a given result). Since $f(z)$ is holomorphic it is continuous and therefore bounded on $\compconj {K_r(0)}$. Using Cauchy–Hadamard theorem it follow from the boundedness that $$\limsup_{n\to\infty} \sqrt[n]{\vert f(z^n)\vert}=1$$
For me, it looks like this is a working proof. But acutally I only used that $f(z)$ is continuous and not that it is holomorphic, what is a way stronger assumption. So can anyone spot my mistake here in this proof? Or is continuity sufficient?

A power series is a series $\sum_{n=0}^\infty a_n z^n$ with coefficients $a_n \in \Bbb C$. The coefficients can not depend on $z$. So you cannot interpret $\sum_{n=1}^{\infty}z^{n} f(z^{n})$ as a power series with coefficients $f(z^{n})$.
But it is correct that the boundedness of $f$ on $\overline{K_r(0)}$ implies the uniform convergence of
$$ \tag{*}
\sum_{n=1}^{\infty}g(z^n)=\sum_{n=1}^{\infty}z^{n} f(z^{n})
$$
$\overline{K_r(0)}$, one can for example use the Weierstrass M-test.
And yes, that would  be true if $f$ were only continuous. In that case one can only conclude that $(*)$ converges to a continuous function.
The holomorphy of $f$ is needed to conclude that the series $(*)$ converges to a holomorphic function (as a locally uniformly convergent series of holomorphic functions).
