I would like to know: How come that

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

Why isn't it infinity?

  • $\begingroup$ See this. (To see that it converges, for $|x|<1$, you could simply use the Ratio Test.) $\endgroup$ Jan 22, 2014 at 15:35
  • $\begingroup$ ... and use $n=(n+1)-1)$. $\endgroup$
    – Rasmus
    Jan 22, 2014 at 15:40

2 Answers 2


Your identity is valid iff $|x|\lt1$. So now assume $|x|<1$ then it holds

$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$

and the convergence is absolute. Hence

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

Multiply both sides with $x$ and you will get

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

But as the first summand for $n=0$ is zero this is the same as $$\sum_{n=1}^\infty nx^{n}=\frac{x}{\left(1-x\right)^2}$$

For $|x|\ge1$ the limit of $nx^n$ does not tend to zero, thus the series $\sum_{n=1}^\infty nx^{n}$ cannot converge in this case.

  • $\begingroup$ It must be mentioned here why the interchange of summation and differentiation is justified. It is justified by the fact that convergent power series converge uniformly on compact subsets of the set of points where they converge. $\endgroup$
    – rayu
    Aug 3, 2018 at 12:33

Using the ratio test, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)x^{n+1}}{nx^n} = \dfrac{(n+1)x}{n}$. So, the series converges when $|x|<1$.

$$F(x) = \sum_{n=1}^\infty n x^n = x + 2x^2 + 3x^3 + ...$$

$$xF(x) = \sum_{n=1}^\infty n x^{n+1} = x^2 + 2x^3 + 3x^4 + ...$$

$$F(x) - xF(x) = x + x^2 + x^3 + x^4... = \sum_{n=1}^\infty x^n = \dfrac{x}{1 - x}$$

It's a geometric series, defined when $|x| < 1$.

$$F(x)(1 - x) = \dfrac{x}{1-x}$$

$$F(x) = \dfrac{x}{(1-x)^2}$$


Not the answer you're looking for? Browse other questions tagged .