Summing a exponential series What is the appropriate way to simplify such an expression.
i am unsure of how to use the series i know to apply to this situation
$$\sum_{L=0}^{M}s^{L}L^{2}$$
do i modify such a series as power series, or is there a more efficient series to use here?
thank you very much!!
 A: Write : $L^2=L(L-1)+L$  and  use  derivative. For $ L \geq 2$ :
$$L^2 s^L = s^2L(L-1)s^{L-2}+ s Ls^{L-1}= s^2(s^L)'' + s (s^L)'$$
We get :
$$\sum_{L=0}^M L^2s^L=0^2+1^2 s + s^2 \left(\sum_{L=2}^{M} s^L \right)''+s \left(\sum_{L=2}^{M} s^L \right)'$$
A: Try to make the inner expression look like a derivative:
$$
\begin{align}
\sum_{L=0}^M\left(Ls^{L-1}\right)sL & =s\sum_{L=0}^M\left(\partial_ss^L\right)L\\
& =s\partial_s\sum_{L=0}^Ms^LL\\
& =s\partial_s\sum_{L=0}^M\left(Ls^{L-1}\right)s\\
& =s\partial_s\left(s\sum_{L=0}^M\left(Ls^{L-1}\right)\right)\\
& =s\partial_s\left(s\sum_{L=0}^M\partial_ss^L\right)\\
& =s\partial_s\left(s\partial_s\sum_{L=0}^Ms^L\right)\\
& =s\partial_s\left(s\partial_s\frac{s^{M+1}-1}{s-1}\right)\\
\end{align}$$
Now just take it from here, simplifying from the inside out.
A: Rewrite $L^2 = L(L-1)+L. $ Then,
$$\begin{align}
\sum_{L=0}^{M} { L^2 s^L } &= \sum_{L=0}^{M} {L(L-1)s^L} + \sum_{L=0}^{M} {Ls^L}\\
&=s^2 \cdot \frac{\partial^2}{\partial s^2} \sum_{L=0}^{M}{s^L} + s \cdot \frac{\partial}{\partial s}\sum_{L=0}^{M} {s^L}\\
\end{align}$$
You can find formulas for the summations with detailed descriptions of their derivations by searching for "Geometric Progression."
