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
\begin{equation}
S = \frac{\ln(10/9) – \psi_{1/10}(1)}{\ln(10)}\, ,
\end{equation}
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.
Best Answer
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 $\sum_{n\geq 1}\frac{d(n)}{10^n}$.