# $\frac{1}{9} + \frac{1}{99} + \frac{1}{999} + \frac{1}{9999} + \cdots ={}$?

Sum the following: \begin{align} S &= \frac{1}{9} + \frac{1}{99} + \frac{1}{999} + \frac{1}{9999} + \cdots\\[0.1in] &= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1} \end{align}

It's fairly straightforward to show that this sum converges: \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1}\\ &= \frac{1}{9} + \sum_{n=2}^{\infty} \frac{1}{{10}^n - 1}\\ &< \frac{1}{9} + \sum_{n=2}^{\infty} \frac{1}{{10}^n - 10}\\ &= \frac{1}{9} + \frac{1}{10}\sum_{n=2}^{\infty} \frac{1}{{10}^{n-1} - 1}\\ &= \frac{1}{9} + \frac{1}{10}S\, , \end{align} which leads to $S < 10/81 = 0.\overline{123456790}$.

Numerically (Mathematica), we find that this sum is approximately $S \approx 0.122324$.

This sum can also be written as \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1}\\ &= \sum_{n=1}^{\infty} \frac{1/{10}^n}{1 - 1/{10}^n}\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{{10}^{nm}} \end{align}

(The last expression above is itself quite amusing, since the coefficient of $1/{10}^k$ is the number of distinct ways of writing $k$ as a product of two positive integers.)

Analytically, Mathematica evaluates this sum as $$S = \frac{\ln(10/9) - \psi_{1/10}(1)}{\ln(10)}\, ,$$ where $\psi_q(z)$ is the Q-Polygamma function. This is not really a nice, tidy, closed-form solution. Now, I can accept that this sum may not have such a nice, tidy sum, but something about it feels like it should have one. (Non-rigorous, I know!) Furthermore, I'm aware that Mathematica is by no means infallible, especially when it comes to simplification of certain expressions.

So I'm wondering if a neater solution exists.

• Wolfram returns the same result. Perhaps if you find a closed form for $\psi_{\frac{1}{10}}(1)$, if that is even possible, then you can have your nice closed-form solution. But I doubt such a closed form exists for $\psi_{\frac{1}{10}}(1)$. – Frpzzd Jun 17 '17 at 15:30
• If you write decimal expansions of the terms (e.g., $\frac19 = .1111111...$; or $\frac1{99} = .010101...$), you can see that the sum is equal to $\sum_1^\infty \frac{d(n)}{10^n}$ where $d(n)$ is the divisor counting function. – paw88789 Jun 17 '17 at 15:57
• @paw88789 That is already stated by the OP in the question. – WimC Jun 17 '17 at 16:27
• As was foreseeable, this type of series has a name and this is about all what one can say. (Why the avalanche of upvotes?) – Did Jun 17 '17 at 18:02
• The smallest number with more than $9$ divisors is $48=2^4\cdot3^1$, so the first 46 digits of the sum are simply the number of divisors of the digit position. – robjohn Jun 17 '17 at 19:07

The given series is an irrational number (probably a trascendental number, too) and for its numerical evaluation it is possible to exploit some tailor-made acceleration techniques (like the one outlined here for the base-$2$ analogue). Besides that, I am not aware of any nice closed form for $\sum_{n\geq 1}\frac{d(n)}{10^n}$.