I have been asked to find the sum to infinity for the following: $$\sum_{i=1}^{\infty}\frac{i}{2^i-1}$$

I attempted to create some sort of relation between successive terms, in order to use a telescoping method, however, I was unsuccessful in doing so. Then I tried to get to a relation between terms $a_i$, $a_{2i}$. However, this too did not help me calculate the required sum.

Any help with this would be greatly appreciated.

  • 1
    $\begingroup$ This is similar to the Erdős–Borwein constant. I am not sure if there is a closed form. $\endgroup$
    – Gary
    Commented Sep 2, 2023 at 7:39
  • $\begingroup$ @Gary. There is an acceptable approximation. $\endgroup$ Commented Sep 2, 2023 at 7:58
  • $\begingroup$ It was not needed to split the sum to arrive to the result in terms of the q-digamma function (just the series). $\endgroup$ Commented Sep 2, 2023 at 9:32

3 Answers 3


Using Abel-Plana -formula:

$$\sum _{x=1}^{\infty } g(x)=-\underset{x\to 0}{\text{lim}}\frac{g(x)}{2}+\int_0^{\infty } g(x) \, dx+i \int_0^{\infty } \frac{g(i x)-g(-i x)}{\exp (2 \pi x)-1} \, dx$$

We have:

$$\sum _{x=1}^{\infty } \frac{x}{2^x-1}=\\-\underset{x\to 0}{\text{lim}}\frac{1}{2}\frac{x}{2^x-1}+\int_0^{\infty } \frac{x}{2^x-1} \, dx+i \int_0^{\infty } \frac{\frac{i x}{-1+2^{-i x}}+\frac{i x}{-1+2^{i x}}}{-1+e^{2 \pi x}} \, dx=\\\color{red}{\frac{1}{24}+\frac{\pi ^2}{6 \ln ^2(2)}-\frac{1}{2 \ln (2)}}$$

And here comes a strange thing, the result is correct with only 22 correct digits !


From here :

$$\sum _{x=1}^{\infty } \frac{x}{a^x-1}=\frac{\psi _{\frac{1}{a}}^{(1)}(1)}{\ln ^2(a)}$$ for: $a>1$

And from here we have:

$$\sum _{x=1}^{\infty } \frac{x}{2^x-1}=\frac{\psi _{\frac{1}{2}}^{(1)}(1)}{\log ^2(2)}=\frac{1}{24}-\frac{E(m) K(m)}{\pi ^2}+\frac{5 K(m)^2}{6 \pi ^2}-\frac{m K(m)^2}{6 \pi ^2}=\color{red}{\frac{-\pi \vartheta _2\left(\frac{1}{2}\right){}^4+5 \pi \vartheta _3\left(\frac{1}{2}\right){}^4-12 \vartheta _3\left(\frac{1}{2}\right){}^2 E\left(\frac{\vartheta _2\left(\frac{1}{2}\right){}^4}{\vartheta _3\left(\frac{1}{2}\right){}^4}\right)+\pi }{24 \pi }}$$ Where: $m=q^{-1}\left(\frac{1}{2}\right)$ is InverseEllipticNomeQ gives the parameter m corresponding to the nome q in an elliptic function.

$K(m)$ is complete elliptic integral of the first kind ,

$E(m)$ is elliptic integral of the second kind

$\psi _{\frac{1}{2}}^{(1)}(1)$ is q-digamma function.

$\vartheta _3\left(\frac{1}{2}\right){}^2$ is theta function.

$\vartheta _2\left(\frac{1}{2}\right){}^4$ is theta function.

$E\left(\frac{\vartheta _2\left(\frac{1}{2}\right){}^4}{\vartheta _3\left(\frac{1}{2}\right){}^4}\right)=-\frac{\pi \text{EllipticThetaPrime}^{(0,1,0)}\left(2,0,\frac{1}{2}\right)}{2 \vartheta _2\left(0,\frac{1}{2}\right) \vartheta _3\left(0,\frac{1}{2}\right){}^2}$

Mathematica code:

Inactive[Sum][x/(-1 + 2^x), {x, 1, Infinity}] == (-Log[2] + QPolyGamma[1, 1, 2])/Log[2]^2 == 1/24 - (EllipticE[InverseEllipticNomeQ[1/2]]*EllipticK[InverseEllipticNomeQ[1/2]])/Pi^2 + (5*EllipticK[InverseEllipticNomeQ[1/2]]^2)/(6*Pi^2) - (EllipticK[InverseEllipticNomeQ[1/2]]^2*InverseEllipticNomeQ[1/2])/(6*Pi^2) == (Pi - Pi*EllipticTheta[2, 1/2]^4 - 12*EllipticE[EllipticTheta[2, 1/2]^4/EllipticTheta[3, 1/2]^4]*EllipticTheta[3, 1/2]^2 + 5*Pi*EllipticTheta[3, 1/2]^4)/(24*Pi) == (1 - EllipticTheta[2, 1/2]^4 + 5*EllipticTheta[3, 1/2]^4 + (6*Derivative[0, 1, 0][EllipticThetaPrime][2, 0, 1/2])/EllipticTheta[2, 1/2])/24

  • 7
    $\begingroup$ It must have to do with the fact that your function is not holomorphic for $\operatorname{Re}(x)\ge 0$. It is beautiful approximation nevertheless. $\endgroup$
    – Gary
    Commented Sep 2, 2023 at 11:51
  • 1
    $\begingroup$ This is more than nice and interesting ! $\endgroup$ Commented Sep 2, 2023 at 17:26
  • 2
    $\begingroup$ Very elegant approximation! A rough estimation gives the value of the error at $x=\frac{2\pi i}{\ln2}$ (where the function gets the first additional singularity and, therefore, loses its analyticity): $$\sim\frac{2\pi}{\ln2}e^{-4\pi^2/\ln2}\approx 1.7\cdot {10}^{-24}$$ $\endgroup$
    – Svyatoslav
    Commented Sep 3, 2023 at 3:38

Writing $$\frac{i}{2^i-1}=\sum_{n=0}^\infty i \,2^{-i (n+1)}$$ $$\sum_{i=1}^\infty i \,2^{-i (n+1)}=\frac{2^{n+1}}{\left(2^{n+1}-1\right)^2}$$ The summation would converge quite fast since $$a_n=\frac{2^{n+1}}{\left(2^{n+1}-1\right)^2}\quad \implies \frac 29<\quad \frac{a_{n+1}}{a_n}< \frac 12$$

$$\sum_{n=0}^\infty \frac{2^{n+1}}{\left(2^{n+1}-1\right)^2}=\frac{\psi _2^{(1)}(1)-\log (2)}{\log ^2(2)}$$ where appears the q-polygamma function.

The summation would converge quite fast since $$a_n=\frac{2^{n+1}}{\left(2^{n+1}-1\right)^2}\quad \implies \frac 29<\quad \frac{a_{n+1}}{a_n}< \frac 12$$


$$\psi _2^{(1)}(1)=2.011526532708124150542153\cdots$$ According to inverse symbolic calculators $$\psi _2^{(1)}(1)\sim \frac{\sqrt[4]{2}+\sqrt[4]{3}\,\, \sqrt[3]{5}}{\sqrt[3]{5}}$$ for an absolute error of $1.998\times 10^{-8}$.

Then, a quite good approximation of the result.

More generally

$$\sum_{i=1}^\infty \frac{i}{\alpha^i-1}=\frac{\psi _{\alpha }^{(1)}(1)-\log (\alpha )}{\log ^2(\alpha )}$$

$$a_n=\frac{\alpha^{n+1}}{\left(\alpha^{n+1}-1\right)^2}\quad \implies \frac{\alpha (\alpha +1)^2}{\left(\alpha ^2+\alpha +1\right)^2}<\quad \frac{a_{n+1}}{a_n}< \frac 1 \alpha$$


Using Euler-MacLaurin summation, we should obtain (with $t=\log(2)$)

$$\frac{\pi ^2}{12 t^2}+\frac{11}{12}+\frac{t}{6}+\frac{t^2}{40}-\frac{13 t^3}{360}-\frac{25 t^4}{1008}+\frac{541 t^5}{15120}+\frac{1561 t^6}{28800}-\frac{47293 t^7}{604800}$$ which is $$2.744033888806\cdots$$ while $$\frac{\psi _2^{(1)}(1)-\log (2)}{\log ^2(2)}=2.74403388876\cdots$$

  • $\begingroup$ Connected : the formula given by Marco Cantarini here, which, by differentiation, gives back your type of result. $\endgroup$
    – Jean Marie
    Commented Sep 2, 2023 at 10:48
  • 1
    $\begingroup$ There appears to be an error with $$\sum_{i=1}^{\infty}\frac{i}{2^i-1}=\frac{\psi _2^{(1)}(1)-\log (2)}{\log ^2(2)} \, .$$ Are you deriving with respect to $q$ instead of $z$ somehow? $\endgroup$
    – Diger
    Commented Sep 2, 2023 at 12:06
  • 1
    $\begingroup$ I'm not familiar with $q$-Gamma function, but from mathworld.wolfram.com/q-PolygammaFunction.html I found $$\psi_q(z)=-\ln(1-q)+\ln(q)\sum_{n=0}^\infty \frac{q^{n+z}}{1-q^{n+z}}$$ and $$\psi'_2(1)=\ln^22\sum_{n=0}^\infty \frac{2^{n+1}}{(1-2^{n+1})^2}$$ which gives $$\frac{\psi'_2(1)-\ln2}{\ln^22} = 1.3013388.... \, .$$ $\endgroup$
    – Diger
    Commented Sep 2, 2023 at 12:17
  • 1
    $\begingroup$ @Diger. The sum from $i=1$ to $i=5$ is, as written , $ \frac{8213}{3255}=2.5232$ $\endgroup$ Commented Sep 2, 2023 at 12:21
  • 1
    $\begingroup$ I'm not sure what you are trying to say, but assuming $\psi_q(z)$ as in my above comment, $$\sum_{i=1}^{\infty}\frac{i}{2^i-1}=\frac{\psi _2^{(1)}(1)-\log (2)}{\log ^2(2)}$$ should be incorrect, but anway... $\endgroup$
    – Diger
    Commented Sep 2, 2023 at 12:32

Not an answer.

Only to give few involved formulas. It can be shown from the theory of elliptic functions that: $$\sum_{n=1}^{\infty}\frac{2n\left(e^{2\pi\sqrt{z}n}+ze^{2\pi n/\sqrt{z}}-z-1 \right)}{(1-e^{2\pi n/\sqrt{z}})(e^{2\pi\sqrt{z}n}-1)}=-\frac{z}{12}+\frac{\sqrt{z}}{2\pi}-\frac{1}{12}.\tag{1}$$ And from this the relation: $$\frac{\pi^2}{12\log^2{2}}\sum_{n=1}^{\infty}\frac{24n}{e^{2\pi^2 n/\log{2}}-1}+\sum_{n=1}^{\infty}\frac{2n}{2^{2n}-1}=-\frac{1}{2\log{2}}+\frac{\pi^2}{12\log^2{2}}+\frac{1}{12}.\tag{2}$$


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