# Summing $\sum_{n=0}^{\infty} n x^n$ [duplicate]

I am trying to figure out how to do the infinite summation:

$$\sum_{n=0}^{\infty} n x^n \qquad 0 \leq x < 1$$

The series converges so it seems to me that the limit must exist, but I'm having difficulties trying to find an exact answer for it. It is very much like an infinite geometric series, but the multiplication factor $n$ makes that you can't use the usual trick. The furthest I came was writing it out as:

$$x + 2 x^2 + 3 x^3 + 4 x^4 + \dots$$

but that doesn't get me anywhere. Can someone give me a hint?

## marked as duplicate by Git Gud, Daniel Fischer, Namaste, Tunk-Fey, hardmathSep 4 '14 at 14:01

• Yes you're right, that question is exactly what I needed. – Yellow Sep 4 '14 at 13:31

This can be written as $$x\sum_{n=1}^{\infty}nx^{n-1}$$ and the series is the derivative of $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

• Though I have seen this many times in my lifetime (and a lot on MSE) this is still up there as a beautifully simple way to solve a rather hard problem :) +1. – Chinny84 Sep 4 '14 at 13:58

Let

$$S = \sum_{n=0}^\infty nx^n$$

Then we know that $$S':= \sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

So $$S + S' = \sum_{n=0}^\infty (n+1)x^n = \frac{S}{x}$$

Therefore $$S = \frac{x}{1-x}S' = \frac{x}{(1-x)^2}$$

For $|x|<1$ we have $\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$ now by derivative we have$\sum\limits_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}=\sum\limits_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$ and so $$\sum\limits_{n=0}^{\infty}nx^{n}=\frac{x}{(1-x)^2}$$