Array of terms or double sum?

Consider the following Maclaurin series for $-1<x<1$ :

$\begin{aligned} \frac 1{1-x} & = 1 + x + x^2 + x^3 + \cdots & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \frac 1{(1-x)^2} & = 1 + 2x + 3x^2 + 4x^3 + \cdots & \small \color{#3D99F6} \text{Multiply both sides by }x^2 \\ \frac {x^2}{(1-x)^2} & = x^2 + 2x^3 + 3x^4 + 4x^5 + \cdots \end{aligned}$

Putting $x = \dfrac 1k$ , we have:

$\begin{aligned} \frac 1{k^2} + \frac 2{k^3} + \frac 3{k^4} + \cdots & = \left(\frac {\frac 1k}{1-\frac 1k}\right)^2 = \boxed{\dfrac 1{(k-1)^2}}\end{aligned}$ .

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Bonus :

$\begin{aligned} S_n & = \sum_{j=n}^\infty \frac j{k^{j+1}} \\ & = \sum_{j=1}^\infty \frac j{k^{j+1}} - \color{#3D99F6} \sum_{j=1}^{n-1} \frac j{k^{j+1}} \\ & = \frac 1{(k-1)^2} - \color{#3D99F6} x^2 \frac d{dx} \left(\frac {x^n-1}{x-1} \right) & \small \color{#3D99F6} \text{where }x = \frac 1k \\ &= \frac 1{(k-1)^2} - x^2\left(\frac {(n-1)x^n-nx^{n-1}+1}{(x-1)^2} \right) \\ &= \frac 1{(k-1)^2} - \frac {k^n-n(k-1)-1}{k^n(k-1)^2} \\ &= \frac 1{(k-1)^2} - \frac 1{(k-1)^2} + \frac {n(k-1)+1}{k^n(k-1)^2} \\ &= \boxed{\dfrac {n(k-1)+1}{k^n(k-1)^2}} \end{aligned}$

Note that when $n=1$ , $S_1 = \dfrac 1{(k-1)^2}$ as expected.