Is this the "standard" proof that $ \sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$? Let $a>0$.  For $|\frac{1}{a^x}|<1, \sum_{n=0}^{\infty} \frac{n\cdot log(a)}{a^{nx}}=\frac{d}{dx}\sum_{n=1}^{\infty} \frac{-1}{a^{nx}}=\frac{d}{dx}\frac{a^x}{1-a^x}=\frac{a^x\cdot log(a)}{(1-a^x)^2}$.  Hence $ \sum_{n=1}^{\infty} \frac{n}{a^{nx}}=\frac{a^x}{(1-a^x)^2}$, which gives $ \sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$, for $|x|<1$.
I don't know how to justify taking the derivative since it's not a power series, but it seems to give me the right answer.  I was wondering if this was the standard way for evaluating this series, or if there is another, more common way, possibly involving generating functions and a differential equation?  
 A: Using derivatives with power series is a useful way to evaluate these power series, when the value of $x$ lies in the interval of convergence. In the case of the series in question, the interval of convergence is $(-1,1)$. So when $x\in (-1,1)$ we have that $\sum\frac{d}{dx}=\frac{d}{dx}\sum$ (i.e. you can swap differentiating the series with taking the series of the differentiated terms).
For this series one could also do the following (for $x\in(-1,1)$):
$$\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=x\sum_{n=1}^\infty\left(\frac{d}{dx}x^n\right)=x\frac{d}{dx}\left(\sum_{n=1}^\infty x^n\right)=x\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{x}{(1-x)^2}$$
A: Usually people use a simpler approach, $x\frac{d}{dx}\sum_{n=1}^\infty x^n$. As with your method, a series that is a function of $x$ is differentiated termwise with respect to $x$. As long as the original and differentiated series have finite terms and converge, this is legitimate even if the differentiated terms aren't a "standard" series expression for a function, such as a Taylor series in $x$.
