Been trying to figure out how to sum $\sum_{k=0}^{\infty} k^{2}x^{k}$ for x<1. I tried the trick through differentiation, but cannot include (k-1) to have second derivative of x. I googled it can be classified as a polylogarithm -2nd order, but I wasnt able to make it work for the sum.
My teacher says the result is $x\cdot \frac{1+x}{(1-x)^{3}}$ wolfram alpha says its the same thing but negative. Any clues how to work it out?
Many thanks
You will need to apply a theorem regarding the differentiation of power series, but on the interval $(-1,1)$ you have $$ \frac{1}{1-x} = \sum_{k=0}^\infty x^k$$ so that $$\frac{1}{(1-x)^2} = \sum_{k=1}^\infty k x^{k-1}$$ and now multiply by $x$ to get $$\frac{x}{(1-x)^2} = \sum_{k=1}^\infty k x^k.$$
Now differentiate and multiply by $x$ one more time.