# Calculate infinite series $\sum_{i=0}^\infty \frac{i^2}{6^i}$

I am asked to calculate the series $$\sum_{i=0}^\infty \frac{i^2}{6^i}$$ From Wolfram Alpha, I know the answer is $\frac{42}{125}$, but I don't know the steps to get the answer. I was told to use $\frac{1}{ 1-a}$.

Can someone help?

• Yes. Derive once and twice the function $6/(6-x) = \sum_1^\infty x^n/6^n$ (which is defined for $|x| < 6$) and put in $x=1$ in the end. – Friedrich Philipp Mar 9 '16 at 0:46

Here is a useful finite evaluation: $$1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1$$ Then by differentiating $(1)$ you get $$1+2r+3r^2+\cdots+nr^{n-1}=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1, \tag2$$ multiplying $(2)$ by $r$ and differentiating gives $$1+4r+9r^2+\cdots+n^2r^{n-1}=\frac{1+r-r^n \left(1+2 n+n^2+r-2 n r-2 n^2 r+n^2 r^2\right)}{(1-r)^3}. \tag3$$
By making $n \to +\infty$ in $(3)$, using $|r|<1$ and multyplying by $r$ we get
$$\sum_{n=1}^\infty n^2r^n=\frac{r(1+r)}{(1-r)^3},\quad |r|<1. \tag4$$
Put $r=\dfrac16$ in $(4)$ to obtain the sought result.