Value of a power series How can I prove for $|x|<1$ the following equality?
$$\frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1}$$
I tried to write
$\sum_{k=1}^\infty kx^{k-1} = (\sum_{k=0}^\infty k+1) (\sum_{k=0}^\infty x^{k})$ = $(\sum_{k=0}^\infty k+1)(\frac{1}{1-x}) $
But the row $\sum_{k=0}^\infty k+1$ diverges. Can someone help me?
Thx
 A: The way you tried to solve it is not mathematical correct.
$$
\sum_{k=1}^{\infty}kx^{k-1} = \frac{d}{dx}\sum_{k=1}^{\infty}x^{k} =\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}
$$
Using the principle of a geometric series.
As pointed out by @Botond, technically the infinite sum should start from $k=0$ but is a constant with respect to $x$ so is ignored as follows
$$
\sum_{k=0}^\infty \frac{d}{dx} x^k =    \frac{d}{dx} x^0 + \sum_{k=1}^\infty \frac{d}{dx} x^k = 0 +\sum_{k=1}^\infty \frac{d}{dx} x^k  
$$
A: 
Differentiation theorem for power series : A power series can be differentiated term-by-term within the interval of convergence. In fact, if $$f(x)=\sum_{n=0}^{\infty} a_n x^n, \; \text {then} \; f'(x)=\sum_{n=1}^{\infty} na_n x^{n-1} \; \text {for} \; |x| \lt R.$$ Both series have same radius of convergence $R$. (From Introduction to Real Analysis by Bartle and Sherbert.)

Can you take it from here? You have to find out $f(x)$ whose derivative is $\frac {1}{1-x^2}$ and which is sum of a well-known geometric series on $|x| \lt 1$.
