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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?"

Did
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user148928
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2 Answers2

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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 $576$.

Did
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  • Well-answered, but one minor caveat: it could also happen that e.g. $d(n)=10$ and $10\leq d(n+1)\lt 20$, or $d(n)=9$ and $20\leq d(n+1)\lt 30$. It's easy to check that neither of these things happens within that range, but there's a 'close call' at $n=495$ where we have two consecutive values with $d(n)\geq 10$ (but the wrong one is equal to 10). – Steven Stadnicki Jul 24 '14 at 21:58
  • @StevenStadnicki Indeed, manually checking that there is no hit before place 576 is necessary to fully tighten the proof (the answer now mentions this fact). In the same vein, one might add a small argument showing that large values of d(n) cannot cascade across more than one or two places before digit 1000, say. – Did Jul 25 '14 at 08:17
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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.

Felix Marin
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