# Sums of infinite series involving the Riemann zeta function

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## 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

1. 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 $$n\rightarrow \infty$$.

2. That formula holds of course, too, if we replace 2m by m; a fact we need in the proof of Theorem 3.2.

3. As usual, the empty sum is defined to be the number 0.

## References

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## Acknowledgements

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

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Correspondence to Raymond Mortini.

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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