Prove that $\sum_{n=0}^{\infty}a_{n}b_{n}z^{n}=\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{f(w)}{w}g(\frac{z}{w})dw$. I'm trying to solve this problem: Let $\rho_{1},\rho_{2}>0$ be the
radius of convergence for the power series $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
and $g(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$, respectively. Prove that
the series $h(z)=\sum_{n=0}^{\infty}a_{n}b_{n}z^{n}$ has radius of
convergence greater than or equal to $\rho_{1}\rho_{2}$. Prove also
that if $|z|<r\rho_{2}$ with $0<r<\rho_{1},$then 
$$
h(z)=\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{f(w)}{w}g\left(\frac{z}{w}\right)dw,
$$
 where $\gamma_{r}$ is the curve defined by $\gamma_{r}(t)=re^{it}$
for $t\in[0,2\pi]$.
I've tried to do this for the first part: Let $\rho$ be the radius
of convergence of the series $h(z)=\sum_{n=0}^{\infty}a_{n}b_{n}z^{n}$.
Then, we know that $\frac{1}{\rho}=\limsup|a_{n}b_{n}|^{\frac{1}{n}}$,
and using the fact that for sequences $(x_{n})_{n\in\mathbb{N}},(y_{n})_{n\in\mathbb{N}}$
is true the inequality $\limsup x_{n}y_{n}\leq\limsup x_{n}\cdot\limsup y_{n}$,
we obtain that:
\begin{align*}
\frac{1}{\rho} & =\limsup|a_{n}b_{n}|^{\frac{1}{n}}\\
 & =\limsup|a_{n}|^{\frac{1}{n}}|b_{n}|^{\frac{1}{n}}\\
 & \leq\limsup|a_{n}|^{\frac{1}{n}}\cdot\limsup|b_{n}|^{\frac{1}{n}}\\
 & =\frac{1}{\rho_{1}}\cdot\frac{1}{\rho_{2}}.
\end{align*}
So $\frac{1}{\rho}\leq\frac{1}{\rho_{1}}\cdot\frac{1}{\rho_{2}}$
and therefore, $\rho\geq\rho_{1}\rho_{2}$.
I don't know if the procedure for this part is completely correct
and also I couldn't prove the part of the integral.
I've tried using Cauchy's integral formula but I haven't obtained
the result. Could you help me or give me some suggestion?
Thanks.
 A: First part looks ok. For the second part, use the following.
\begin{align*}
\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{f(w)}{w}g\left(\frac{z}{w}\right)\,dw&=\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{1}{w}\sum_{n=0}^{\infty}a_{n}w^{n}\sum_{m=0}^{\infty}b_{m}\left(\frac{z^{m}}{w^{m}}\right)\,dw
\\
&=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}z^{m}\frac{1}{2\pi i}\int_{\gamma_{r}}w^{n-m-1}\,dw.
\end{align*}
Interchanging of summations and integral is allowed because $\gamma_{r}(t)=re^{it}$ for $0<r<\rho_{1}$.
When $n=m$, the integral
\begin{equation*}
\frac{1}{2\pi i}\int_{\gamma_{r}}w^{n-m-1}\,dw=1
\end{equation*}
by parametrizing $w=re^{it}$. When $n>m$, equivalently, $n\geq m+1$ or $n-m-1\geq 0$, the integral,
\begin{equation*}
\frac{1}{2\pi i}\int_{\gamma_{r}}w^{n-m-1}\,dw=0
\end{equation*}
by Cauchy-Goursat Theorem.
When $n<m$, equivalently, $n+1\leq m$ or $n-m-1\leq -2$, let $k=n-m-1\leq -2$, the integral
\begin{equation*}
\frac{1}{2\pi i}\int_{\gamma_{r}}w^{n-m-1}\,dw=0
\end{equation*}
by Cauchy's Integral Formula for higher derivatives. Therefore,
\begin{equation*}
\frac{1}{2\pi i}\int_{\gamma_{r}}\frac{f(w)}{w}g\left(\frac{z}{w}\right)\,dw=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}z^{m}\frac{1}{2\pi i}\int_{\gamma_{r}}w^{n-m-1}\,dw=\sum_{k=0}^{\infty}a_{k}b_{k}z^{k}=h(z).
\end{equation*}
