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How to sum this series:

$$\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$$

My attempt:

Multiply and divide the series by $9$

$$9\left(\frac{1}{9}+\frac{1}{99}+\frac{1}{999}+\frac{1}{9999}+\cdots\right)$$

$$9\left(\frac{1}{10-1}+\frac{1}{10^2-1}+\frac{1}{10^3-1}+\frac{1}{10^4-1}+\cdots\right)$$

Now let $a_N$ denote the number of divisors of $N$, after some simplification the series becomes:

$$9\left(1+\sum{\frac{a_N}{10^N}}\right)$$

This is where I am stuck...

PS: Please rectify my mistakes along the way

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  • $\begingroup$ Are you sure about the problem? W|A gives answer in terms of q-digamma function: wolframalpha.com/input/… $\endgroup$ May 16, 2014 at 13:18
  • $\begingroup$ For reference, this is an example of a Lambert series. $\endgroup$
    – David H
    May 16, 2014 at 13:33
  • $\begingroup$ And as an aside, the analogous infinite series using powers of $2$ instead of powers of $10$, i.e. $E=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$, is known as the Erdős-Borwein constant. $\endgroup$
    – David H
    May 16, 2014 at 13:40
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    $\begingroup$ There must be some relationship to the primes. The addition of the decimal fraction features a pattern similar to the sieve of Erastothenes. At all p-th position, where $p$ is a prime, the $9$ fits in just right, where $p$ is not a prime we have additions of several $9$s. $\endgroup$
    – mvw
    May 16, 2014 at 13:53

1 Answer 1

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Your approach is very nice but, as pointed by Pranav Arora, for the summation up to term $n$, a CAS leads to $$S =9 \left(\frac{\psi _{\frac{1}{10}}^{(0)}(n+1)}{\log (10)}-\frac{\psi _{\frac{1}{10}}^{(0)}(1)}{\log (10)}\right)$$ and for the infinite summation, it becomes $$S=\frac{9 \left(\log \left(\frac{10}{9}\right)-\psi _{\frac{1}{10}}^{(0)}(1)\right)}{\log (10)} \simeq 1.100918190836200736379855$$

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