Sum of a power series $n x^n$ [duplicate]

Question

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?

Answer 1

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

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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.

Answer 2

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 $|x| < 1$.

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

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