TITLE: Calculating the series $1/8+1/88+1/888+....$
QUESTION [7 upvotes]: I wonder whether this series is calculable or not.
Attempt:
$S=1/8+1/88+1/888+....=\dfrac18\displaystyle\sum_{k=0}^\infty\dfrac{1}{\sum_{n=0}^k10^n}$

where $$\displaystyle\sum_{n=0}^k10^n=\dfrac{10^{k+1}-1}{9}$$

then
$S=\dfrac98\displaystyle\sum_{k=0}^\infty\dfrac{1}{10^{k+1}-1}$
I have tried to calculate $\displaystyle\sum_{k=0}^K\dfrac{1}{10^{k+1}-1}$ for finite values but I failed.
What methods can we try?

REPLY [3 votes]: $$\frac{1}{8}+\frac{1}{88}+\frac{1}{888}+\dotsm = \frac{1}{8}\left( 1+\frac{1}{11}+\frac{1}{111}+\dotsm  \right) =
 \frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \iff x= \frac{1}{10}$$
Where $\sum_{n=1}^{\infty} \frac{x^n}{1-x^n}$ Is the Lambert series for the sequence given by $a_n = 1$
For this specific case we have:
$$
 S=\frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} =\frac{9}{8}\left(\frac{\log\left(1-x\right)+\psi_{x}^{(0)}(1)}{log(x)}\right)
$$ 
That gives us:
$$S=\frac{9}{8}\left(\frac{\log\left(\frac{9}{10}\right)+\psi_{\frac{1}{10}}^{(0)}(1)}{log(\frac{1}{10})}\right)=0.137614773854525092047481887706797505400431...$$
Where $\psi_{x}^{(y)}(z)$ is the generalized PolyGamma function