# $1+ {1\over 11}+ {1\over 111}+ {1\over 1111}+....=?$ [closed]

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.

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

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.

• "Closed form" is debatable here since the psi-function is defined precisely as the sum of this series.
– Did
Commented Jan 12, 2019 at 16:25
• @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. Commented Jan 12, 2019 at 16:44