sequences and series - proof of $sum_{n=1}^infty n cdot x^n= frac{x}{(x-1)^2}$

I know that the Series $\sum_{n=1}^\infty n \cdot x^n$ converges to $\frac{x}{(x-1)^2}$ but I'm not sure how to show it. I'm pretty sure that has been asked before, but I wasn't able to find anything...

Answer

Let $$S_n=\sum_{k=1}^n kx^k$$

Assuming $|x|<1$,
$$S_n-xS_n=\sum_{k=1}^n x^k-nx^{n+1}=\frac{x(1-x^n)}{1-x}-nx^{n+1}\Rightarrow S_n=\frac{x(1-x^n)}{(1-x)^2}-\frac{nx^{n+1}}{1-x}$$

Now, if $|x|<1$, then $$\limsup_{n\rightarrow \infty}\left|\frac{(n+1)x^{n+1}}{nx^n}\right|=|x|\limsup_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)=|x|<1$$ So, by ratio test, the series $\displaystyle \sum_{k=1}^n kx^k$ converges and hence $$\lim_{n\rightarrow \infty}nx^{n+1}=x\lim_{n\rightarrow \infty}nx^{n}=0$$

So, $$S=\sum_{n=1}^{\infty}nx^n=\lim_{n\rightarrow \infty}S_n=\frac{x}{(1-x)^2}$$