# Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$

Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$

My approach :

$= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$

Now how to solve this I am not getting any clue on this please help thanks.

## 2 Answers

Starting with \begin{align} \frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n} \end{align} then differentiate and multiply by $t$ to obtain \begin{align} \frac{t}{(1-t)^{2}} = \sum_{n=0}^{\infty} n t^{n}. \end{align} Repeating leads to \begin{align} \sum_{n=0}^{\infty} n^{2} \ t^{n} = \frac{t(1+t)}{(1-t)^{3}}. \end{align} Now let $t = 1/6$ for which \begin{align} \sum_{n=0}^{\infty} \frac{n^{2}}{6^{n}} = \frac{42}{125}. \end{align}

• Oh right, this is $Li_{-2}(6^{-1})$. Completely missed that. – Gina Jun 10 '14 at 4:08

Break $n^2$ into two parts: $\underbrace{n(n-1)} + \underbrace{{}\quad n\quad {}}$.

The first part appears in the second derivative of $x^n$ and the second part in the first derivative:

\begin{align} \sum_{n=1}^\infty n^2 x^n & = \sum_{n=1}^\infty x^2\Big(n(n-1) x^{n-2} \Big) + x\Big(nx^{n-1}\Big) \\[10pt] & = \left(x^2 \frac{d^2}{dx^2} \sum_{n=1}^\infty x^n\right) + \left( x\frac{d}{dx}\sum_{n=1}^\infty x^n \right) \\[10pt] & = \left( x^2 \frac{d^2}{dx^2} \frac{x}{1-x}\right) + \left( x\frac{d}{dx} \frac{x}{1-x} \right) \\[10pt] & = \cdots\cdots \end{align}

(Finally at the end, put $x=1/6$.)