How to show that for $|x|<1$, $1+2x+3x^2+\cdots=\frac1{(1-x)^2}$? I saw this formula and I don't know if it has a certain name or how can it be proven

If $|x|<1$ then $$1+2x+3x^2+4x^3+\cdots= \frac{1}{(1-x)^2}$$

Can someone tell me anything about it? How can it be proven?
 A: This is a power series: $$\sum_{n=0}^\infty(n+1) x^n$$
For power series we can integrate term by term and we have:
$$\int\left(\sum_{n=0}^\infty(n+1) x^n\right)dx = \sum_{n=0}^\infty\int (n+1)x^ndx = \sum_{n=0}^\infty x^{n+1} = \sum_{n=1}^\infty x^n = \frac x{1-x}$$
So if we differentiate $\frac{x}{1-x}$ with respect to $x$, and we get the sum of the original series:
$$\left(\frac x{1-x}\right)'=\frac1{1-x}+\frac{x}{(1-x)^2}=\frac1{(1-x)^2}$$
As wanted. Of course we have to have $|x|<1$ for everything to converge and have meaning.
A: If you know that
$$
1+x+x^2+x^3 + \cdots = \frac{1}{1-x}
$$
then you can get this identity by differentiating both sides.
This raises a problem: Is the derivative of the sum equal to the sum of the derivatives even when there are infinitely many terms?  In fact, in some cases it is not.  However, it can be shown that for a power series (i.e. each term is a number times a different power of $x$) that does work.
That, however, is not the only way to do it.
A: Here is a solution which doesn't use integration or differentiation of power series.
I am going to use 
$$1+x+x^2+x^3 + \cdots +x^k= \frac{1-x^{k+1}}{1-x}$$
is well known and easy to prove. I will assume that you know it, but if you don't I can provide a proof.
Let 
$$S_k=1+2x+3x^2+..+(k+1)x^k $$
$$T_k=1+x+x^2+..+x^k$$
Then
$$S_k-T_k=x+2x^2+3x^3+..+kx^k=xS_{k-1}=x(S_k-(k+1)x^k )$$
Thus
$$S_k(x-1)=(k+1)x^{k+1}-T_k=(k+1)x^{k+1}-\frac{1-x^{k+1}}{1-x}$$
Thus
$$S_k= \frac{1}{(1-x)^2}+\frac{(k+1)x^{k+1}}{x-1}-\frac{x^{k+1}}{(x-1)^2}$$
Since $\lim_k (k+1)x^{k+1}=\lim_k x^{k+1} =0 \, \forall |x|<1$ you get the desired result.
