Value of $x=\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}} $ and location of second digit $1$ of $x$ 
What is the value of the series
  $\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}}$
  and can one find closed forms for the sums
  $\sum\limits_{n=1}^N {\frac{1}{10^n-1}}$?

I found that
$\sum_{n=1}^N {\frac{1}{10^n-1}} $  =  $\sum_{n=1}^N (\lim_{m\to\infty  }{\frac{10^{-n}-10^{-nm}}{1-10^{-n}}})$  =  $0.111\ldots+0.0101\ldots+\ldots$
but I can't go farther.

The original question is: "Where does the second '1' appear?"

 A: Expanding the fraction $\frac{x}{1-x}$ as the series $\sum\limits_{k\geqslant1}x^k$ for each $x=\frac1{10^n}$ yields $$\sum_{n\geqslant1}\frac1{10^n-1}=\sum_{n\geqslant1}\frac{\frac1{10^n}}{1-\frac1{10^n}}=\sum_{n,k\geqslant1}\frac1{10^{kn}}.$$
In the RHS, each $\frac1{10^i}$ appears as many times as one can decompose the integer $i\geqslant1$ as $i=kn$, that is, $d(i)$ times, where $d(i)$ denotes the number of divisors of $i$, also denoted $\sigma_0(i)$. Thus,
$$\sum_{n\geqslant1}\frac1{10^n-1}=\sum_{i\geqslant1}\frac{d(i)}{10^i}=\frac1{10}+\frac2{10^2}+\frac2{10^3}+\frac3{10^4}+\frac2{10^5}+\frac4{10^6}+\cdots,$$
that is,
$$
\sum_{n\geqslant1}\frac1{10^n-1}=0.1223242434262445\ldots$$
Note that $d(n)\geqslant2$ for every $n\geqslant2$. But the digit $1$ can also appear at place $n$ when $d(n+1)\geqslant10$ has a unit digit $1$ and when $d(n+2)\leqslant9$. According to this OEIS table of the sequence $(d(n))$, this happens for the first time with $d(576)=21$ and $d(577)=2$, hence, after manually checking there is no lower hit, this proves that the second digit $1$ is at place $576$.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
\color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over 10^{n} - 1}}&=
\sum_{n = 0}^{\infty}{\pars{1/10}^{n + 1} \over 1 - \pars{1/10}^{n + 1}}
={\Psi_{\rm{1/10}}\pars{1} + \ln\pars{1 - 1/10} \over \ln\pars{1/10}}
\\[3mm]&=\color{#66f}{\large1 - {\ln\pars{9} + \Psi_{\rm{1/10}}\pars{1} \over \ln\pars{10}}}\approx {\tt 0.1223}
\end{align}

where $\Psi_{\rm q}\pars{z}$ is the q-PolyGamma Function and we used identity $\ds{\pars{2}}$ in that link.

