I've had a test in probability, during which I've had to find the convergence value of
$\sum\limits_{i=1}^\infty n^2 \dot (x)^{n-1} $ when $x=\frac{5}{6}$.
I couldn't find the answer during the test.
That sum expression was part of calculating an expected value of a random variable, that I know that had one, so I don't need to prove first that the series converge.
Wolfram Alpha tells me the answer is $\sum\limits_{i=1}^\infty n^2 \dot (x)^{n-1}= -\frac {x+1} {(x-1)^3}$ when $|x| < 1$, but I don't know how it got to it.
Any explanation on how to find this series' convergence value will be greatly appreciated.
Observe that $$ \sum_{n = 1}^{\infty} n^2 x^{n-1} = x \sum_{n=2}^{\infty} n(n-1) x^{n-2} + \sum_{n = 1}^{\infty} n x^{n-1} $$ You can obtain formulae for $\sum_{n\geq 1} n x^{n-1}$ and $\sum_{n \geq 2} n (n-1) x^{n-2}$ by differentiating $(1-x)^{-1} = \sum_{n=0}^{\infty} x^n$ once and twice resepctively.