Prove infinite series $$
\frac{1}{x}+\frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \cdots =\frac{x}{(x-1)^2}
$$
I can feel it. I can't prove it. I have tested it, and it seems to work. Domain-wise, I think it might be $x>1$, the question doesn't specify. Putting the LHS into Wolfram Alpha doesn't generate the RHS (it times out).

I think a less formal solution could be more understandable.
consider $$  S_n= \frac{1}{x} + \frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \dots + \frac{n}{x^n}$$
$$ xS_n = 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \dots + \frac{n}{x^{n-1}}$$
then
$$xS_n - S_n = 1+ (\frac{2}{x}-\frac{1}{x})+(\frac{3}{x^2}-\frac{2}{x^2})+\dots+(\frac{n}{x^{n-1}}-\frac{n-1}{x^{n-1}}) - \frac{n}{x^n}$$
$$S_n(x-1) = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}} - \frac{n}{x^n}$$
Now we have a simplified the problem to one of a basic geometric series, so
$$S_n(x-1) = T_{n-1} - \frac{n}{x^n}$$
where
$$T_{n-1} = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}}$$
$$\frac{T_{n-1}}{x} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+\dots+\frac{1}{x^{n}}$$
$$T_{n-1} - \frac{T_{n-1}}{x} = 1 + (\frac{1}{x}-\frac{1}{x})+ (\frac{1}{x^2}-\frac{1}{x^2})+\dots - \frac{1}{x^n}$$
$$T_{n-1}(1-\frac{1}{x}) = 1  - \frac{1}{x^n}$$
$$T_{n-1}(\frac{x-1}{x}) = \frac{x^n-1}{x^n}$$
$$T_{n-1} = \frac{x^n-1}{x^n}\cdot(\frac{x}{x-1})$$
$$T_{n-1} = \frac{x^n-1}{x-1}\cdot(\frac{1}{x^{n-1}})$$
$$T_{n-1} = \frac{x-\frac{1}{x^{n-1}}}{x-1}$$
Thus $S_n(x-1)$ becomes
$$S_n(x-1) = \frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$
for $|x|\gt 0$ this becomes
$$\lim_{n\to\infty}S_n(x-1) = \lim_{n\to\infty}\frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$
$$S(x-1) = \frac{x-\displaystyle\lim_{n\to\infty}\frac{1}{x^{n-1}}}{x-1} - \lim_{n\to\infty}\frac{n}{x^n}$$
$$S(x-1) = \frac{x-0}{x-1} - 0 = \frac{x}{x-1}$$
$$S =  \frac{x}{(x-1)^2}$$
I used l'Hopital's rule to evaluate $\displaystyle\lim_{n\to\infty}\frac{n}{x^n}$, being an $\frac{\infty}{\infty}$ indeterminate form
This helps me to understand the problem. Afterwards, I would go on to compose a more formal proof.
