<s>[INST] 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.
 [/INST] You have this
$(z - 1)S 
= 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}
$
or, in summation notation,
$(z - 1)S 
= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \frac{i^2}{z^i}
$.
We can now split this into sums
we already know:
$\begin{array}\\
(z - 1)S 
&= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \dfrac{i^2}{z^i}\\
&= \sum_{k=0}^{i-1} \dfrac{2k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\
&= 2\sum_{k=0}^{i-1} \dfrac{k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\
\end{array}
$
You can now plug in
the known summations.
More generally,
if
$S_m(z)
=\sum_{k=0}^{\infty} \dfrac{k^m}{z^k}
$,
then
$S_0(z)
=\sum_{k=0}^{\infty} \dfrac{1}{z^k}
=\dfrac{1}{1-1/z}
=\dfrac{z}{z-1}
$
and,
for $m \ge 1$,
$S_m(z)
=\sum_{k=1}^{\infty} \dfrac{k^m}{z^k}
$,
$zS_m(z)
=\sum_{k=1}^{\infty} \dfrac{k^m}{z^{k-1}}
=\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}}
$
so
$\begin{array}\\
(z-1)S_m(z)
&= zS_m(z)-S_m(z)\\
&=\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}}-\sum_{k=0}^{\infty} \dfrac{k^m}{z^k}\\
&=\sum_{k=0}^{\infty} \dfrac{(k+1)^m-k^m}{z^{k}}\\
&=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^m \binom{m}{j}k^j-k^m}{z^{k}}\\
&=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^{m-1} \binom{m}{j}k^j}{z^{k}}\\
&=\sum_{j=0}^{m-1} \binom{m}{j}\sum_{k=0}^{\infty} \dfrac{k^j}{z^{k}}\\
&=\sum_{j=0}^{m-1} \binom{m}{j}S_j(z)\\
\end{array}
$
so
$S_m(z)
=\dfrac1{z-1}\sum_{j=0}^{m-1} \binom{m}{j}S_j(z)
$
so that each
$S_m(z)
$
can be gotten in terms
of the
$S_j(z)
$
for $j < m$.
In your case,
the $2$ and the $1$ comes from
$(k+1)^2-k^2
=2k+1
$.
</s>