Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer

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): schematic

share|cite|improve this answer
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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.