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 first encountered this sum with the following problem:
Evaluate
$$\bigg(\frac{1}{2}\bigg)^\dfrac{1}{3}\bigg(\frac{1}{4}\bigg)^\dfrac{1}{9}\bigg(\frac{1}{8}\bigg)^\dfrac{1}{27}\bigg(\frac{1}{16}\bigg)^\dfrac{1}{81}\dots$$

Which , of course  simplified to 
$$\bigg(\frac{1}{2}\bigg)^{\dfrac{1}{3^1}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+\dots}=\bigg(\frac{1}{2}\bigg)^S$$
Getting back to your problem, now
$$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}=\frac{2}{3}\sum_{n=1}^\infty \frac{n}{3^n}=\frac{2}{3}S$$
Using a method similar to deriving geometric series suppose that
$$S_k = \sum_{n=1}^k \frac{n}{3^{n}}$$
Then  we have 
$$\begin{array}{lll}
3S_k-S_k &=& 1+\frac{2}{3^1}+\frac{3}{3^2}+\frac{4}{3^3}+\dots+\frac{k}{3^{k-1}}\\
&&-\frac{1}{3^1}-\frac{2}{3^2}-\frac{3}{3^3}\dots-\frac{k-1}{3^{k-1}}-\frac{k}{3^k}\\
2S_k&=&\bigg(1+\frac{2-1}{3^1}+\frac{3-2}{3^2}+\frac{4-3}{3^3}+\dots+\frac{k-(k-1)}{3^{k-1}}\bigg)-\frac{k}{3^k}\\
&=&\frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\
2S&=&\lim_{k\to\infty} \frac{1-(\frac{1}{3})^{k}}{1-\frac{1}{3}} - \frac{k}{3^k}\\
2S&=&\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\\
\frac{2}{3}S&=&\frac{1}{2}\\
\end{array}$$
by a similar method one can show, that if the series converges, that 
$$\sum_{n=0}^\infty (n+1)x^n = \frac{1}{(1-x)^2}$$
