How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$

I assume that the $|x|$ to be less than $1$. Now, consider,
$f(x)=\sum_{n=0}^{n=\infty} x^{n+1}$ 
This will converge only if $|x|<1$. Now, interesting thing here is, this is a geometric progression. The $f(x)=x/(1-x)$.
$f'(x)$ is the series you are interested in, right? Differentiate $x/(1-x)$ and you have your expression! 
