Summing a exponential series

Question

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!!

Answer 1

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.

Answer 2

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 : $$\sum_{L=0}^M L^2s^L=0^2+1^2 s + s^2 \left(\sum_{L=2}^{M} s^L \right)''+s \left(\sum_{L=2}^{M} s^L \right)'$$

Answer 3

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."