There is No Backward

$\begin{aligned} S & = \sum_{k=1}^\infty \frac k{5^k} \\ & = \frac 15 \sum_{k=1}^\infty \frac k{5^{k-1}} \\ & = \frac 15 \sum_{k=1}^\infty \frac {\partial \color{#3D99F6}{x}^k}{\partial \color{#3D99F6}{x}} & \small \color{#3D99F6}{\text{Let }x = \frac 15} \\ & = \frac 15 \times \frac {\partial}{\partial x} \sum_{k=1}^\infty x^k \\ & = \frac 15 \times \frac {\partial}{\partial x} \left(\frac x{1-x} \right) \\ & = \frac 15 \left( \frac 1{1-x} + \frac x{(1-x)^2} \right) \\ & = \frac 15 \times \frac 1{(1-x)^2} & \small \color{#3D99F6}{\text{Putting back }x = \frac 15} \\ & = \frac 5{16} = \boxed{0.3125} \end{aligned}$