A problem by Romeo Gomez

I will do it for general cases. For $k\in\mathbb{N}-\{1\}$ , let define the sum as $x=\sum_{n=1}^{\infty}{\frac{n}{k^n}=\frac{1}{k}+\frac{2}{k^2}+\frac{3}{k^3}+\dots},$ divide $x$ by $k$ $\frac{x}{k}=\sum_{n=1}^{\infty}{\frac{n}{k^{n+1}}}=\frac{1}{k^2}+\frac{2}{k^3}+\frac{3}{k^4}+\dots,$ then subtract it to the original sum $x-\frac{x}{k}=\frac{1}{k}+(\frac{2}{k^2}-\frac{1}{k^2})+(\frac{3}{k^3}-\frac{2}{k^3})+\dots$ $x(\frac{k-1}{k})=\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+\dots$ $x(\frac{k-1}{k})=\sum_{n=1}^{\infty}{\frac{1}{k^n}}$ $x(\frac{k-1}{k})=\sum_{n=1}^{\infty}{\frac{1}{k^n}}+1-1$ $x(\frac{k-1}{k})=\sum_{n=0}^{\infty}{\frac{1}{k^n}}-1,$ we get a Geometric serie $x(\frac{k-1}{k})=\frac{1}{1-\frac{1}{k}}-1$ $x(\frac{k-1}{k})=\frac{1}{\frac{k-1}{k}}-1$ $x(\frac{k-1}{k})=\frac{k}{k-1}-1$ $x(\frac{k-1}{k})=\frac{1}{k-1}$ $x=\frac{k}{(k-1)^2},$ so $\sum_{n=1}^{\infty}{\frac{n}{k^n}}=\frac{k}{(k-1)^2},$ let $k=2015$ $\sum_{n=1}^{\infty}{\frac{n}{2015^n}}=\frac{2015}{2014^2}$ so $Romeo=\boxed{4058211},$ yey we find him !!!!!!! yujuuuu!!!! :D