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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?

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  • $\begingroup$ by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test $\endgroup$ – pHotone Aug 1 '18 at 8:07
  • $\begingroup$ I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series? $\endgroup$ – pHotone Aug 1 '18 at 8:11
  • $\begingroup$ WolframAlpha doesn't give an elementary answer, so chances are there isn't one. There are simplifications that WA misses, though, so we can't be certain. $\endgroup$ – Arthur Aug 1 '18 at 8:23
  • $\begingroup$ This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function" $\endgroup$ – Turing Aug 1 '18 at 8:23
  • $\begingroup$ The sum $$s(z):=\sum_{n=0}^\infty\,\frac{1}{z^n-1}\text{ for }z>1$$ involves the $q$-polygamma function $\psi^{(0)}_{q}$ with $q:=\frac1z$. WolframAlpha gives $$s(z)=\frac{\ln\left(\frac{z}{z-1}\right)-\psi^{(0)}_{\frac1z}(1)}{\ln(z)}\text{ for }z>1\,.$$ See mathworld.wolfram.com/q-PolygammaFunction.html. $\endgroup$ – Batominovski Aug 1 '18 at 8:28
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$$\displaystyle\sum_{k=0}^K\dfrac{1}{10^{k+1}-1}=-1-K+\frac{\psi _{10}^{(0)}(K+2)-\psi _{10}^{(0)}(1)}{\log (10)}$$ where appears the generalized PolyGamma function.

It is not surprising that you have problems with it.

Edit

If $K \to \infty$, the limit is $$S=-\frac{9 }{16 \log (10)}\left(2 \psi _{10}^{(0)}(1)+\log \left(\frac{81}{10}\right)\right)\approx 0.13761477385452509205$$

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$$\frac{1}{8}+\frac{1}{88}+\frac{1}{888}+\dotsm = \frac{1}{8}\left( 1+\frac{1}{11}+\frac{1}{111}+\dotsm \right) = \frac{9}{8}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \iff x= \frac{1}{10}$$

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

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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.

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    $\begingroup$ The sign is wrong? Shouldn't it be $\frac{1}{10^k-1}$ instead? $\endgroup$ – Batominovski Aug 1 '18 at 8:29
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    $\begingroup$ I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+..... $\endgroup$ – pHotone Aug 1 '18 at 14:10

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