Representing $\sum\limits_{n=1}^\infty a_{n-1}a_nz^n$ in terms of $\sum\limits_{n=0}^\infty a_nz^n$ Let $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ be an analytic function in some open set $D$ with $a_n>0$ for all $n\geq 0$. Then there are various series that can be represented in terms of $f$. For example, we have that $\sum\limits_{n=1}^\infty na_nz^n=zf'(z).$  Is there a way to represent the series $\sum\limits_{n=1}^\infty a_{n-1}a_nz^n$ in terms of $f$?
 A: Here's what I got: by Cauchy's generalized integral formula,
$$\begin{split}
g(z) &= \sum_{n=1}^\infty a_{n-1}a_n z^n = \sum_{n=1}^\infty \frac{f^{(n-1)}(0)}{(n-1)!} a_n z^n =  \sum_{n=1}^\infty \frac{a_nz^n}{2\pi i} \int_{\circlearrowleft} \frac{f(\zeta)}{\zeta^n}d\zeta \\
&=\frac{1}{2\pi i} \int_\circlearrowleft \left[f\left( \frac z \zeta \right) - f(0) \right] f(\zeta) d\zeta,
\end{split}$$
where $\circlearrowleft$ is a counterclockwise contour circling the origin. The formula makes sense: since $f$ is defined as a Maclaurin series and $D$ is an open set, $D \subseteq B_r(0)$, where $0 < r \leq \infty$ is the radius of convergence, and if $D \neq B_r(0)$, $f$ can be extended to the open ball $B_r(0)$. It suffices to take $\circlearrowleft$ to be any circumference of radius $< 1$. Is this what you wanted?
A: This is essentially a Hadamard product of $f$ with $zf$: https://en.wikipedia.org/wiki/Generating_function_transformation#Hadamard_products_and_diagonal_generating_functions
It's not a very natural operation, and requires an integral to be expressed in closed form. The expression in the Wikipedia article gives (a bit sloppy with square roots I guess, but never mind!):
$$
(f \odot zf)(z) = \frac{1}{2\pi} \int_0^{2\pi} f(\sqrt{z}e^{it}) f(\sqrt{z}e^{-it}) \sqrt{z}e^{-it} {\rm d}t
$$
or slightly more neatly,
$$
(f \odot zf)(z^2) = \frac{z}{2\pi} \int_0^{2\pi} f(z e^{it}) f( ze^{-it}) e^{-it} {\rm d}t
$$

Example: $f(z) = \exp(z)$ gives
$$\begin{align}
(f \odot zf)(z)
&= \frac{z^{1/2}}{2\pi} \int_0^{2\pi} \exp(\sqrt{z}e^{it}) \exp(\sqrt{z}e^{-it}) e^{-it} {\rm d}t \\ 
&= \frac{z^{1/2}}{2\pi} \int_0^{2\pi} \exp(2\sqrt{z} \cos t - it){\rm d}t \\
&= z^{1/2} I_1(2\sqrt{z})
\end{align}$$
in terms of a particular Bessel function, which indeed has the series
$$z^{1/2} I_1(2\sqrt{z}) = \sum_{n=1}^\infty \frac{z^n}{n!(n-1)!}$$
This example shows that you don't expect a simple expression (e.g. not involving integrals) for the result.

Edit: Note that the intuition behind the formula 
$$(f \odot g)(z) = \frac{1}{2\pi} \int_0^{2\pi} f(\sqrt{z}e^{it}) g(\sqrt{z}e^{-it})  {\rm d}t$$
is that you want a given power of $z^{n}$ in the answer to arise from an equal balance of terms $f_n z^{n/2} \times g_n z^{n/2}$ from $f,g$. Hence you just multiply $f$ and $g$ together, but give the term  $\propto f_n$ a phase factor $e^{i n t}$ and the term $\propto g_n$ a phase factor $e^{-i n t}$ and then look for things which have no overall phase. This is what integrating $\frac{1}{2\pi} \int {\rm d}t$ does.
