# Simple infinite sum identity

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.

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To show that $\sum_{n=1}^{\infty}\frac{n}{10^n}= \frac{10}{81},$ why not just show (by long division) that $\frac{10}{81} = \overline{0.1234567890}$ and then notice that the partial sums of the infinite series are $0.1, 0.12, 0.123, \dots, 0.12345678901, 0.123456789012,\dots?$ Then you can use geometric series identities to show that the nested series have the same sum, which demonstrates the identity. – barf Aug 3 '11 at 0:47

Writing $x$ for $1/10$ we have $$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}\sum_{m=1}^nx^n=\sum_{m=1}^{\infty}\sum_{n=m}^{\infty}x^n=\sum_{m=1}^{\infty}x^m\sum_{n=m}^{\infty}x^{n-m}=\sum_{m=1}^{\infty}x^m\sum_{n=0}^{\infty}x^n$$ and the last sum is your right-hand side (except that my $m$ is your $n$, and vice versa). As Qiaochu notes, absolute convergence is necessary for the interchange-of-summations step.

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Personally, I'd go the following route: $$\sum_{n=1}^\infty\frac{n}{10^n}=\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)+\left(\sum_{n=2}^\infty\frac{1}{10^n}\right)+\left(\sum_{n=3}^\infty\frac{1}{10^n}\right)+\cdots$$ $$=\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)+\frac{1}{10}\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)+\frac{1}{10^2}\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)+\cdots$$ $$=\left(1+\frac{1}{10}+\frac{1}{10^2}+\cdots \right)\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)$$ $$=\left(\sum_{m=0}^\infty\frac{1}{10^m}\right)\left(\sum_{n=1}^\infty\frac{1}{10^n}\right)=\frac{10}{9}\cdot\frac{1}{9}=\frac{10}{81}.$$

But that's just me.

EDIT: I added a fun little schematic to justify the first step to the mind's eye (or whatever):

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The first step is well justified in my mind, but I'm looking for an analytic proof rather than a heuristic one. Or is this already the simplest it gets? – jnm2 Aug 3 '11 at 11:33
@jnm2: in what way is this a heuristic proof? Are you familiar with the fact that an absolutely convergent series can be summed in any order? – Qiaochu Yuan Aug 4 '11 at 17:52
This answer seemed heuristic because the justification was "for the mind's eye" (the summation notation was basically identical to the schematic). I was already aware of the behavior and was looking for a more formal proof. And no, I'm not very familiar with this, so maybe I'm just looking stupid. – jnm2 Aug 5 '11 at 16:35

The RHS can be written $\displaystyle \sum_{n \ge 1, m \ge 0} \frac{1}{10^{n+m}}$. Collect all terms with the same value of $n+m$. To justify this rigorously you just need to know that the series converges absolutely, which should be clear.

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Are you allowed to use derivatives? If so, then...

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$

$\frac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1}$

$\frac{x}{(1-x)^2} = \sum_{n=1}^\infty n x^n$

Letting $x=\frac{1}{10}$, you get

$\frac{10}{81} = \sum_{n=1}^\infty \frac{n}{10^n}$

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