Summing $\sum_{n=0}^{\infty} n x^n$

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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 $$\sum\limits_{n=0}^{\infty}nx^{n}=\frac{x}{(1-x)^2}$$