## Abstract

We determine the values of several infinite series involving the Riemann zeta function. In particular, degree one rationally weighted summands involving the Riemann function evaluated at even numbers give a finite sum involving only the Riemann function evaluated at odd numbers.

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## Notes

Actually \(\lim _{n\rightarrow \infty }(\zeta (2n)-1) 4^{n}=1\). In fact,

$$\begin{aligned} 4^n(\zeta (2n)-1) =\sum _{k=2}^\infty \left( \frac{2}{k}\right) ^{2n}=1 + \sum _{k=3}^\infty \left( \frac{2}{k}\right) ^{2n}. \end{aligned}$$Now the convergence of \(\sum _{k=3}^\infty \left( \frac{2}{k}\right) ^{2x}\) is uniform in \(x\ge 1\) (due to the converging majorant \(\sum _{k=3}^\infty \left( \frac{2}{k}\right) ^{2}\)). Hence \(\lim _{x\rightarrow \infty }\sum _k=\sum _k\lim _{x\rightarrow \infty }\). The assertion follows, as each summand goes to 0 as \(n\rightarrow \infty \).

That formula holds of course, too, if we replace 2

*m*by*m*; a fact we need in the proof of Theorem 3.2.As usual, the empty sum is defined to be the number 0.

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## Acknowledgements

We thank Roberto Tauraso for providing us reference [6].

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Mortini, R., Rupp, R. Sums of infinite series involving the Riemann zeta function.
*Arch. Math.* **123**, 163–172 (2024). https://doi.org/10.1007/s00013-024-02008-7

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DOI: https://doi.org/10.1007/s00013-024-02008-7