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Given the following infinite sum, what methods of attack might yield an answer?$$f(k) = \sum^\infty_{n=1}\frac{1}{k^n-1}$$

I understand there might be some kind of Euler-like transform but I can't find it and even then it seems like another dead end: $$f(k) = \prod_{p\in\mathbb{P}}\frac{?}{?}$$

I've also tried computational methods but don't recognize the results. For example, $$f(2) \approx 1.606695$$ $$f(3) \approx 0.68215$$$$f(4) \approx 0.42109$$

David Diaz
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  • There's no Euler transform since $\frac{1}{k^n-1}$ (as a function of $k$) is not multiplicative – mathworker21 Jun 29 '18 at 02:06
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    An answer to which question? Such series (approximately geometric) are simple to compute numerically, since one might devise tailor-made acceleration techniques (see here). They have been proved to produce trascendental numbers for any $k\geq 2$ by Shingo Takeuchi. – Jack D'Aurizio Jun 29 '18 at 16:19

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$$f(k) = \sum^\infty_{n=1}\frac{1}{k^n-1}=\frac{\log \left(\frac{k}{k-1}\right)-\psi _{\frac{1}{k}}^{(0)}(1)}{\log (k)}$$ where appears the q-digamma function.

For large values of $k$, series expansion gives $$f(k)=\frac{1}{k}+\frac{2}{k^2}+\frac{2}{k^3}+\frac{3}{k^4}+\frac{2}{k^5}+\frac{4}{k^6}+\frac{2}{k^7}+\frac{4}{k^8}+O\left(\frac{1}{k^9}\right)$$ For $k=4$, this would give $\frac{6899}{16384}\approx 0.421082$ which is not too bad compared to the value you gave for $f(4)$.

Claude Leibovici
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