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Can we prove that $$\sum_{n=1}^{\infty}\frac{1}{2^{n}-1}$$ is irrational?

Wolfram Alpha says that it is equal to $$1-\frac{\Psi_{1/2}(1)}{\log 2}$$ where $\Psi_{q}(z)$ is a $q$-polygamma function.

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Yes, the irrationality of $\sum\limits_n\frac1{t^n-1}$ for any integer $|t|>1$ is a famous result of Erdős. The elementary proof may be found here.

See also the Wikipedia page on the Erdős–Borwein constant.

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We have $$ S=\sum_{n\geq 1}\frac{1}{2^n-1}=\sum_{n\geq 1}\frac{d(n)}{2^n}=\frac{5}{4}+\sum_{k\geq 2}\frac{8^k+1}{(2^k-1)\,2^{k^2+k}} \tag{1}$$ hence the aperiodicity of the binary representation of $S$ (from which the irrationality of $S$ follows) can be proved by exploiting the acceleration formula or the fact that $d(n)=\sigma_0(n)$ is odd iff $n$ is a square, and the set of squares does not belong to any arithmetic progression.

It is relevant to mention that Tachiya has proved an interesting generalization of the above claim, by producing acceleration formulas through umbral calculus techniques.

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  • $\begingroup$ Can we use the same argument for $\sum 1/(10^n-1)$? $\endgroup$ – Seewoo Lee Jul 19 '17 at 13:36
  • $\begingroup$ @See-WooLee: yes, there is very little to change, the principle stays the same. $\endgroup$ – Jack D'Aurizio Jul 19 '17 at 13:38
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    $\begingroup$ I'm sorry but I have more questions. Is the series $\sum d(n)/2^n$ gives its binary expansion? I think we need some regrouping when we adding them since $d(n)$ is bigger than 1. Also, I can't understand you first argument.. what is the acceleration formula? $\endgroup$ – Seewoo Lee Jul 19 '17 at 13:54
  • $\begingroup$ Everything is standing just in front of you. The binary representation is given by $\sum_{n\geq 0}\frac{c_n}{2^n}$ with $c_n\in\{0,1\}$, hence the middle series is clearly related with the binary representation. The acceleration formula is the last equality in $(1)$: we convert a series whose general term behaves like $\frac{\log}{2^n}$ into a series whose general term behaves like $\frac{1}{2^{n^2}}$, with a massive improvement about the speed of convergence of partial sums. $\endgroup$ – Jack D'Aurizio Jul 19 '17 at 13:58

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