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What is the sum of the series $$1+ {1\over 11}+ {1\over 111}+ {1\over 1111}+....$$.The partial sum is a monotonically increasing and bounded above sequence, so sum must exits in real.

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    $\begingroup$ Not sure if this helps, but you could also write the sum as $\frac{9}{9}+\frac{9}{99}+\frac{9}{999}+...$, which becomes $\sum_{i=1}^\infty \frac{9}{10^i-1}$. $\endgroup$ Commented Jan 12, 2019 at 15:16
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    $\begingroup$ Did you come up with this question? I'm not sure it convergence to anything special. $\endgroup$
    – Yanko
    Commented Jan 12, 2019 at 15:17
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    $\begingroup$ Wolfram Alpha gave me a closed-form answer to this sum in terms of the $q$-digamma function, but I'm not sure how they derived this answer: wolframalpha.com/input/… $\endgroup$ Commented Jan 12, 2019 at 15:18
  • $\begingroup$ @NobleMushtak wow never heard of this function. $\endgroup$
    – Yanko
    Commented Jan 12, 2019 at 15:19
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    $\begingroup$ In fact, it's a corollary of Eq. (4) here with $a=10$. $\endgroup$
    – J.G.
    Commented Jan 12, 2019 at 15:34

1 Answer 1

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This answer follows on from Noble Mushtak's comments regarding the simplification of the sum, and the closed form solution on Wolfram Alpha.

The q-digamma function can be written as

$$\psi_q(z)=-\ln(1-q)+\ln q\sum_{n=0}^\infty\frac{q^{n+z}}{1-q^{n+z}}$$

So the sum $$\sum_{n=1}^\infty\frac{9}{10^n-1}=9\sum_{n=0}^\infty\frac{10^{-n-1}}{1-10^{-n-1}}$$ So if we let $q=\frac1{10}$, then this is $$9\sum_{n=0}^\infty\frac{q^{n+1}}{1-q^{n+1}}=\frac{9\left(\psi_{\frac1{10}}(1)+\ln\frac9{10}\right)}{\ln\frac1{10}}=\frac{9\left(\ln\frac{10}9-\psi_{\frac1{10}}(1)\right)}{\ln{10}}$$

As given by Wolfram Alpha.

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    $\begingroup$ "Closed form" is debatable here since the psi-function is defined precisely as the sum of this series. $\endgroup$
    – Did
    Commented Jan 12, 2019 at 16:25
  • $\begingroup$ @Did Agreed, I just wanted to be clear that it was the same solution that had been referred to in the comments of the question. $\endgroup$
    – John Doe
    Commented Jan 12, 2019 at 16:44

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