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Given $S = \sum_{n=0}^\infty a_n$ where $a_n$ is $\sum_{i=n}^\infty{x^i}$, we have $S = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots$

I'm trying to find its sum over the reals on the interval $x \in (-1, 1)$.

I can prove that $x\cdot a_n = a_n-x^n$ using the infinite series representation. Therefore $$a_n = \frac{x^n}{1-x} \tag{1}$$.

So here's my work from there.

$$ \begin{align} S &= \sum_{n=0}^\infty{\frac{x^n}{1-x}}\\ \label{metric}&= \frac{\sum x^n}{1-x}\\ &= \frac{\frac1{1-x}}{1-x}\tag{2} \\ &= \frac1{(1-x)^2} \end{align}$$

$2$ uses the same logic as $1$ in summing the terms.

Is this a logical and valid way to find $S$?

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    Should it be $xa_n=a_n-x^n$？Moreover $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$. Thus the final answer is the same as the one you got. – 2017-02-01
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    Ah, I see. I suppose I did that wrong, but luckily it turned out the same as if I had done it another way. Thanks for catching my mistake! – 2017-02-01

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You have absolute convergence when $x \in (-1,1)$ so you can get away with anything on rearranging the sums. Yes, the approach is fine once you show that and the result is correct. You dropped minus signs twice. One is in Frank Lu's comment. The other is that $\sum x^n=\frac 1{1-x}$.