# Finding sum of infinite series

• Sep 10th 2010, 08:43 PM
oleholeh
Finding sum of infinite series
Hi
I'm not sure if it belongs in calculus but here it is anyway:

Find the sum of $\displaystyle \displaystyle\sum_{i=1}^{\infty}\frac{i^2}{a^i}$ for a constant $\displaystyle a$

Wolframalpha yielded $\displaystyle \frac{a(a+1)}{(a-1)^3}$ but I would like to know how to derive it by hand.
• Sep 10th 2010, 08:57 PM
Failure
Quote:

Originally Posted by oleholeh
Hi
I'm not sure if it belongs in calculus but here it is anyway:

Find the sum of $\displaystyle \displaystyle\sum_{i=1}^{\infty}\frac{i^2}{a^i}$ for a constant $\displaystyle a$

Wolframalpha yielded $\displaystyle \frac{a(a+1)}{(a-1)^3}$ but I would like to know how to derive it by hand.

The standard trick is to do it by differentiation, for consider this (assuming $\displaystyle |a|<1$, of course)

$\displaystyle \frac{d}{da}\sum\limits_{i=1}^\infty\left(\frac{1} {a}\right)^i = \sum\limits_{i=1}^\infty\frac{d}{da}\left(\frac{1} {a}\right)^i=\sum\limits_{i=1}^\infty i\left(\frac{1}{a}\right)^{i-1}=\frac{d}{da}\frac{1}{1-a}$

Now, apply this trick twice and you find that

$\displaystyle \sum\limits_{i=1}^\infty \frac{i^2}{a^i}=a\cdot\frac{d}{da}\left(a\cdot \frac{d}{da}\frac{1}{1-a}\right)$