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

Note that $n+1$ is the number ways to choose $n$ items of $2$ types (repetitions allowed but order is ignored), so that $n+1=\left(\!\binom2n\!\right)=(-1)^n\binom{-2}n$. (This uses the notation $\left(\!\binom mn\!\right)$ for the number of ways to choose $n$ items of $m$ types with repetition, a number equal to $\binom{m+n-1}n=(-1)^n\binom{-m}n$ by the usual definiton of binomial coefficients with general upper index.) Now recognise the binomial formula for exponent $-2$ in
$$
  \sum_{n\geq0}(n+1)x^n=\sum_{n\geq0}(-1)^n\tbinom{-2}nx^n
 =\sum_{n\geq0}\tbinom{-2}n(-x)^n=(1-x)^{-2}.
$$
This is valid as formal power series in$~x$, and also gives an identity for convergent power series whenever $|x|<1$.
There is a nice graphic way to understand this identity. The terms of the square of the formal power series $\frac1{1-x}=\sum_{i\geq0}x^i$ can be arranged into an infinite matrix, with at position $(i,j)$ (with $i,j\geq0$) the term$~x^{i+j}$ . Now for given $n$ the terms $x^n$ occur on the $n+1$ positions with $i+j=n$ (an anti-diagonal) and grouping like terms results in the series $\sum_{n\geq0}(n+1)x^n$.
