Calculating the series $1/8+1/88+1/888+....$

Question

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?

by the way I am sure this series is convergent since we can compare it with p series, edit :or just using ratio test — Micheal Brain Hurts, Aug 01 '18 at 08:07
I have seen once a theorem that we cannot sometimes calculate indefinite integrals in forms of elementary functions, are there any theorem for series? — Micheal Brain Hurts, Aug 01 '18 at 08:11
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. — Arthur, Aug 01 '18 at 08:23
This is not an easy series to calculate. In fact we need a basic knowledge of what is called the "Polygamma function" — Enrico M., Aug 01 '18 at 08:23
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 http://mathworld.wolfram.com/q-PolygammaFunction.html. — Batominovski, Aug 01 '18 at 08:28

Claude Leibovici · Accepted Answer · 2018-08-03T07:26:59.743

4

$$\displaystyle\sum_{k=0}^K\dfrac{1}{10^{k+1}-1}=-1-K+\frac{\psi _{10}^{(0)}(K+2)-\psi _{10}^{(0)}(1)}{\log (10)}$$ where appears the generalized PolyGamma function.

It is not surprising that you have problems with it.

Edit

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

edited Aug 03 '18 at 07:26

answered Aug 01 '18 at 08:31

Claude Leibovici

260,315

score 3 · Answer 2 · answered Aug 03 '18 at 10:29

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

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

score 2 · Answer 3 · answered Aug 01 '18 at 08:24

2

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.

answered Aug 01 '18 at 08:24

Enrico M.

26,114

1

The sign is wrong? Shouldn't it be $\frac{1}{10^k-1}$ instead? – Batominovski Aug 01 '18 at 08:29
1

I think Batominovski is right since Von Neumann's series like 1/8(1/11+1/101+1/1001+..... – Micheal Brain Hurts Aug 01 '18 at 14:10

Calculating the series $1/8+1/88+1/888+....$

3 Answers3

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