2014
DOI: 10.1155/2014/457620
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Rapidly Converging Series forζ(2n+1)from Fourier Series

Abstract: Ever since Euler first evaluatedζ(2)andζ(2m), numerous interesting solutions of the problem of evaluating theζ(2m)  (m∈ℕ)have appeared in the mathematical literature. Until now no simple formula analogous to the evaluation ofζ(2m)  (m∈ℕ)is known forζ(2m+1)  (m∈ℕ)or even for any special case such asζ(3). Instead, various rapidly converging series forζ(2m+1)have been developed by many authors. Here, using Fourier series, we aim mainly at presenting a recurrence formula for rapidly converging series forζ(2m+1). I… Show more

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Cited by 7 publications
(9 citation statements)
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References 25 publications
(16 reference statements)
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“…Numerical result show that 20 points are capable of producing an accuracy of seven-decimal place for small arguments. Besides, many rapidly converging series for ζ(2n + 1) have been introduced by Srivastava in a review article [13] and by other authors [12,14].…”
Section: Calculation Of Riemann Zeta Function At Odd Integersmentioning
confidence: 99%
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“…Numerical result show that 20 points are capable of producing an accuracy of seven-decimal place for small arguments. Besides, many rapidly converging series for ζ(2n + 1) have been introduced by Srivastava in a review article [13] and by other authors [12,14].…”
Section: Calculation Of Riemann Zeta Function At Odd Integersmentioning
confidence: 99%
“…where we have used the fact that ζ(s) < , which is superior to other rapid series |R 25 | < 0.9 × 10 −18 as noted in [12,8]. Specially, when N is larger than some typical numbers, the asymptotic behavior of (37) reads (38) lg(|R (s) N |) ∼ −2 lg 2(N + 1) − 3 lg N. The accuracy of the latter two methods rely on the number of zeros(denoted as N 1 ) of the associated polynomial(in this occasion it is Hermite polynomial) and the terms(denoted as N 2 ) of partial sum of the infinite series respectively.…”
Section: Calculation Of Riemann Zeta Function At Odd Integersmentioning
confidence: 99%
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“…which can also be used to evaluate ζ(2n) (n ∈ N \ {1}) by recalling the Basler problem ζ(2) = π 2 /6 (see, e.g., [6] and the references cited therein).…”
Section: Introductionmentioning
confidence: 99%
“…Ever since Euler first evaluated ζ(2) and ζ(2n), numerous interesting solutions of the problem of evaluating the ζ(2n) (n ∈ N) have appeared in the mathematical literature. Even though there were certain earlier works which gave a rather long list of papers and books together with some useful comments on the methods of evaluation of ζ(2) and ζ(2n) (see, e.g., [9], [25], [45] and [75]), the reader may be referred to the very recent work [19] which contains an extensive literature of as many as more than 70 papers. Among many different ways to prove (3), several authors have taken advantage of the nice interplay between a double integral and a geometric series (see, e.g., [5,11,39]).…”
mentioning
confidence: 99%