<s>[INST] Generating function of $\frac{h(x)}{(1-x)^2}$ 
If $h(x)$ is the generating function for $a_r$, what is the generating function of $$\frac{h(x)}{(1-x)^2}$$

Let $h(x)$ be written as
$$h(x) = \sum_{r} a_r x^r $$
Consider more simply
$$\frac{h(x)}{1-x} = \frac{1}{1-x} h(x) =\sum_{r} x^r \sum_{r} a_r x^r$$
I tried to expand this and see what I could get
$$(1+x+x^2+x^3+\dots+x^r+\dots)(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_rx^r+\dots)$$
there are two ways to simplify the product, either
$$a_0(1+x+x^2+\dots)+a_1(x+x^2+x^3+\dots)+ a_2(x^2+x^3+x^4+\dots)+\dots= \sum_ra_r\sum_{k\ge r}x^k$$
or
$$a_0 + (a_0+a_1)x+(a_0+a_1+a_2)x^2+(a_0+a_1+a_2+a_3)x^3 = \sum_r \left(\sum_{k\le r}a_k \right)x^r$$
obviously this is only for one factor of $\frac{1}{1-x}$ but I assume If I can get help for this I can extend it to two factors.
I'm not sure what form the answer is expected to be in? Because I could say the generating function is
$$\frac{h(x)}{1-x} = h\left(\sum_{k \ge r}x^k\right)$$
but I'm not sure that makes any sense. I was expecting to say something like
$$\frac{h(x)}{1-x} \mapsto h(x^2)$$
(the $x^2$ is not intentional, just some idea of what I believe the answer could look like)
Any help for the case of $\frac{1}{1-x}$ would be great and then I could extend it to $\frac{1}{(1-x)^2}$
 [/INST] Another way.
Since
$(1-x)^2 = 1-2x+x^2$,
if
$\dfrac{h(x)}{(1-x)^2}
=\sum_{n=0}^{\infty} a_nx^n$
then
$\begin{array}\\
h(x)
&=(1-x)^2\sum_{n=0}^{\infty} a_nx^n\\
&=(1-2x+x^2)\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n-2x\sum_{n=0}^{\infty} a_nx^n+x^2\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=0}^{\infty} 2a_nx^{n+1}+\sum_{n=0}^{\infty} a_nx^{n+2}\\
&=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=1}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\
&=a_0+a_1x+\sum_{n=}^{\infty} a_nx^n-2a_0x-\sum_{n=2}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\
&=a_0+(a_1-2a_0)x+\sum_{n=}^{\infty} (a_n-2a_{n-1}+a_{n-2})x^n\\
\end{array}a_n
$
If
$h(x)
=\sum_{n=0}^{\infty} h_nx^n$
then,
equating coefficients,
$h_0 = a_0$,
$h_1 = a_1-2a_0$,
so
$a_1 
= h_1+2a_0
= h_1+2h_0
$,
and,
for $n \ge 2$,
$h_n
=a_n-2a_{n-1}+a_{n-2}
$
so
$a_n
=h_n+2a_{n-1}-a_{n-2}
$.
The advantage of this method
is that getting each new $a_n$
takes only
3 operations
(multiply, add, subtract)
instead of $n$.
</s>