Find the sum of series without differentiation Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum
My method does not require differentiation but there is a difficulty.
Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^i}$
Let $zS = 1 + \frac{2^2}{z} + \frac{3^2}{z^2} + ... + \frac{i^2}{z^{i - 1}}$
Thus, $zS - S = 1 + \frac{2^2 - 1^2}{z} + \frac{3^2 - 2^2}{z^2} + ... + \frac{i^2 - (i - 1)^2}{z^{i - 1}} - \frac{i^2}{z^i}$
Thus, $(z - 1)S = 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}$
My question is how can I proceed? The numerator of each term is not 1 so I cannot use any formula to calculate the sum.

Let
\begin{eqnarray*}
S=\sum_{i=1}^{\infty} i^2 x^i.
\end{eqnarray*}
Now multiply by $(1-3x+3x^2-x^3)$ and note that for $i \geq 2$
\begin{eqnarray*}
(i-3)^2-3(i-2)^2+3(i-1)^2-i^3=0.
\end{eqnarray*}
Examine the lower order terms more carefully, and we have
\begin{eqnarray*}
 (1-3x+3x^2-x^3)S=x(1+x)
\end{eqnarray*}
giving the well known formula
\begin{eqnarray*}
S=\sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
