I'm looking for a concise way to show this: $\sum_{n=1}^{\infty}\frac{n}{10^n} = \sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right)$ With this goal in mind: $\sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right) = \sum_{n=1}^{\infty}\left(\left(\frac{10}{9}\right){10^{-n}}\right) = \frac{10}{81}$
So far I've been looking at it by replacing $n$ in the LHS with $(\sum_{m=1}^{n}1)$ like this: $\sum_{n=1}^{\infty}\left(\left(\sum_{m=1}^{n}{1}\right){10^{-n}}\right) = \sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right)$
And here I hit a particularly uncreative brick wall. This equation is obvious to me in a common sense way - I could easily demonstrate it by writing out the RHS as a huge addition problem and showing that the LHS just has the digit columns added ahead of time - but I don't know what to do in between for a proof.