<s>[INST] 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 $$\sum_{n=0}^\infty (n+1)x^n?$$
 [/INST] Note that $\int \{1 + 2x + 3x^2 + \cdots\} \, dx = x + x^2 + x^3 + \cdots + \text{const}$, i.e., a geometric series, which converges to $x/(1 - x)$ if $|x| < 1$. Therefore,
$$\frac{d}{dx} \left(\frac{x}{1 - x}\right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2},$$
that is,
$$1 + 2x + 3x^2 + \cdots = \frac{1}{(1 - x)^2}.$$
Another proof: Let $S = 1 + 2x + 3x^2 + \cdots$ with $|x| < 1$. Then
$$xS = x + 2x^2 + 3x^3 + \cdots$$
so
$$S - xS = (1- x)S = 1 + x + x^2 + \cdots = \frac{1}{1- x}.$$
Therefore: $S = (1 - x)^{-2}$.
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