# How can I evaluate $\sum_{n=1}^{\infty} n^{3}x^{n-1}$

## How do i evaluate :

$$\sum_{n=1}^{\infty} n^{3}x^{n-1}$$

The answer is supposed to be: (according to wolfram alpha)

$$\frac{x^2+4x+1}{(x-1)^4}$$

I have only learned to to this for simpler geometric sums.

Play with these things:

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n\implies x\left(\frac x{(1-x)^2}\right)'=\sum_{n=1}^\infty n^2x^n$$

and etc. Observe the above is valid for $\;|x|<1\;$ ...

So you know the usual geometric entire series, like:

$$\sum_{n \geqslant 1} x^n = \frac{1}{1-x}$$

on its convergence domain. By the good analytic properties of entire series, you can derivate term by term, obtaining:

$$\sum_{n \geqslant 1} nx^{n-1} = \frac{1}{(1-x)^2}$$

And if you derivate three times, you will get your awaited $n^3$ appearing, giving you the $(1-x)^4$. Just write it properly in order to determine the constants (take care when derivating more than once), and work a little to get exactly $n^3x^{n-1}$, you will naturally obtain something slightly different.

Hint: Note $$\sum \limits_{k=0}^{\infty} x^k = \frac{1}{1-x}$$ for $|x|<1$. Now differentiate and multiply this identity with $x$ (and argue why you may interchange the limit and the derivative). After doing this multiple times, you get the result.

• For $|x|<1$ actually. – principal-ideal-domain Dec 12 '16 at 14:15
• @principal-ideal-domain You are right, thanks. In the first version of my post I wrote a finite partial sum, that's where $x \ne 1$ came from. – Dominik Dec 12 '16 at 14:18