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 $$\sum_{n=0}^\infty (n+1)x^n?$$

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$ 

Notice that 
\begin{align*}
S_{m}-rS_{m} & = -mr^{m+1}+\sum_{n=1}^{m}r^{n}\\
  & = -mr^{m+1}+\frac{r-r^{m+1}}{1-r} \\
& =\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.
\end{align*}
Hence 
$$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum.    That is 
\begin{align*}
\sum_{n=1}^{\infty}\frac{2n}{3^{n+1}} 
& = \frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}} \\
& =\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}} \\
& =\frac{1}{2}.
\end{align*}

Added note:  

We can define $$S_m^k(r) = \sum_{n=1}^m n^k r^n.$$  Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$.  We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$.  This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.  

This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$?  It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^{k+1}$.
