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).
 A: For $\;|x|<1\;$ :
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies \frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies$$
$$\implies\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n$$
So no: you don't have an equality there...
Added on request: If $\;|x|>1\;$ then we can do:
$$\frac x{(x-1)^2}=\frac1x\frac1{\left(1-\frac1x\right)^2}=\frac1x\left(\sum_{n=0}^\infty\frac1{x^n}\right)^2$$
So still not the same expression as in the question.
A: Let's put this together cleanly:
For the sum to converge, $|x| > 1$ which means
 $$\sum_{n=0}^\infty x^{-n} = \frac{x}{x-1}$$
Squaring this gives
 $$\frac{x^2}{(x-1)^2} = \sum_{n=0}^\infty \sum_{m=0}^\infty x^{-(n+m)} $$
$$ = \sum_{r=0}^\infty (r+1)x^{-r}$$
(because each value of $(n+m)$ occurs $(n+m+1)$ times in the infinite sum (0 = 0+0, 1 = 1+0 or 0+1, 2=0+2 or 1+1 or 2+0, and so on).
Divide by $x$ gives:
$$ \frac{x}{(x-1)^2} = \sum_{r=0}^\infty (r+1)x^{-(r+1)} $$
$$ = \frac1{x} + \frac{2}{x^2} + \frac{3}{x^3} + ... $$
A: Consider
$$
\sum_{n=0}^\infty y^n=\frac{1}{1-y}\quad;\quad\text{for}\ |y|<1.\tag1
$$
Differentiating $(1)$ with respect to $y$ yields
$$
\sum_{n=1}^\infty ny^{n-1}=\frac{1}{(1-y)^2}.\tag2
$$
Multiplying $(2)$ by $y$ yields
$$
\sum_{n=1}^\infty ny^{n}=\frac{y}{(1-y)^2}.\tag3
$$
Now plug in $y=\dfrac1x$ where $|x|>1$ to $(3)$ yields
$$
\large\color{blue}{\sum_{n=1}^\infty \frac{n}{x^n}=\frac{x}{(x-1)^2}}.
$$
A: HINT: for $|y|<1$
$$\sum_{n=0}^{\infty} n y^n=y\cdot\frac{d \sum_{n=0}^{\infty} y^n}{dy}=y\cdot\frac{d\left(\dfrac1{1-y}\right)}{dy}=\frac y{(1-y)^2}$$
A: 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.
