Can anyone show and explain how to obtain the derivative of the following function, $$y(x)= \sum_{n=1}^\infty nx^{-n}$$
which is, for $n=3$, equivalent with,
$$y(x)=x^{-1}+2x^{-2}+3x^{-3}$$
now, what is the derivative of $x$ with respect to $y$, i.e. $\dfrac{dx}{dy}$ for $n=\infty$?
Can this kind of problem be solved analytically?
Hint. Your function is defined for $|x|>1$ (the series is convergent for such $x$) and, see for example Solve $\sum nx^n$, $$y(x)= \sum_{n=1}^\infty nx^{-n}=\frac{1/x}{(1-1/x)^2}=\frac{x}{(1-x)^2}.$$ Therefore $yx^2-(2y+1)x+y=0$. Can you take it from here?
For $|x|>1$ , let us put
$$y (x)=z (\frac 1x) $$ with
$$z (x)=\sum_{n=1}^{+\infty}nx^n $$
$$=x+2x^2+3x^3+...$$ $$=x (1+2x+3x^3+...)=x\frac {d (x+x^2+x^3+...)}{dx} $$ $$=x\frac {d\frac {x}{1-x}}{dx} =\frac {x}{(1-x)^2} $$
but $$y'(x)=\frac {-1}{x^2}z'(\frac 1x) $$
with $$z'(x)=\frac {1+x}{(1-x)^3} $$
we find that $$y'(x)=z' (x) $$
Let's see:
$$ \sum\limits_{n=1}^{\infty} n x^{-n} = x \sum\limits_{n=1}^{\infty} n x^{n-1} = x \sum\limits_{n=1}^{\infty} \frac{d}{d x}x^{n} = x \frac{d}{dx}\sum\limits_{n=1}^{\infty} x^{n} = x \frac{d}{dx}\left( \frac{1}{1-x} - 1 \right) = \frac{x}{(1-x)^2} $$
where the second to last equality is from the geometric series formula (assuming $|x| < 1$).
From this Solve $\sum nx^n$ See that for $|x|<1$ we have, $$y\left(\frac{1}{x}\right) = \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$$
Hence $$y\left(x\right) = \frac{x}{(1-x)^2},~~~~~|x|>1$$ can you conclude?
Maybe your question is about the derivation of $y_N(x)=\sum_{n=1}^N n\cdot x^{-n}$?
then you can say that, fixed N, $y'_N(x)=-\sum_{n=1}^N n^2\cdot x^{-n-1}$, naturally valid in the domain of the function $y_N$. The problem is when you want to calculate the derivation of $\sum_{n=1}^\infty n \cdot x^{-n}$. Then you have to use some Criterions of convergence of series in order to derivate term by term.
