Calculating the series $1/8+1/88+1/888+....$ 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?
 A: $$\displaystyle\sum_{k=0}^K\dfrac{1}{10^{k+1}-1}=-1-K+\frac{\psi _{10}^{(0)}(K+2)-\psi _{10}^{(0)}(1)}{\log (10)}$$ where appears the generalized  PolyGamma function.
It is not surprising that you have problems with it.
Edit
If $K \to \infty$, the limit is
$$S=-\frac{9 }{16
   \log (10)}\left(2 \psi _{10}^{(0)}(1)+\log \left(\frac{81}{10}\right)\right)\approx 0.13761477385452509205$$
A: $$\frac{1}{8}+\frac{1}{88}+\frac{1}{888}+\dotsm = \frac{1}{8}\left( 1+\frac{1}{11}+\frac{1}{111}+\dotsm  \right) =
 \frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \iff x= \frac{1}{10}$$
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
A: I will extend the answer soon, but in the meanwhile I throw the rock and I give you the direct answer:
$$\frac{1}{8} \sum_{k = 0}^{+\infty} \frac{1}{10^k + 1} = \frac{1}{8}\frac{-\log \left(\frac{10}{9}\right)+\psi _{\frac{1}{10}}^{(0)}\left(-\frac{i \pi }{\log (10)}\right)}{\log (10)}$$
Where $\psi _{\frac{1}{10}}^{(0)}$ is the above mentioned PolyGamma generalized function.
