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In the textbook https://www.springer.com/gp/book/9783034851688 (Klassische elementare Analysis, by M. Koecher) the following elegant recurrence relation is proved for $\zeta(2n)$ (on p. 157): $$\left(n+\frac{1}{2}\right)\zeta(2n)=\sum\limits_{m=1}^{n-1}\zeta(2m)\,\zeta(2n-2m). \tag{1}$$
In fact (1) is equivalent to Euler's recurrence relation for Bernoulli numbers (independently found by Ramanujan) $$(2n+1)B_{2n}=-\sum\limits_{m=1}^{n-1}\binom{2n}{2m}B_{2m}\,B_{2n-2m}. \tag{2}$$ Why, In contrast to (2), (1) can seldom be found in the literature (I was able to find only https://link.springer.com/article/10.1007/s00591-007-0022-2 that mentions (1))? Are there any other references that discuss (1)?

P.S. In addition to juan's answer. G.T. Williams was not the first to state the result in this form. It can be found at least in N. Nielsen, Handbuch der theorie der gammafunktion, Leipzig: Druck und Verlag von B.G. Teubner, 1906, p. 49. I found this reference thanks to the paper "Some identities involving the Riemann zeta function. II." by R. Sitaramachandrarao and B. Davis, Indian J. Pure Appl. Math. 17(10):1175–1186, 1986. https://www.insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20005a50_1175.pdf This reference also has (1) and proves (among others) an interesting generalization of (1): $$4\sum\limits_{i+j+k=n}\zeta(2i)\zeta(2j)\zeta(2k)=(n+1)(2n+1)\zeta(2n)-6\zeta(2)\zeta(2n-2),$$ where $n\ge 3$ and the sum extends over all ordered triples $(i,j,k)$ of positive integers satisfying $i+j+k=n$.

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I have this recurrence in my collection of problems for my lessons in Analytic Number Theory. I have there the reference:

P. Ribenboim, Classical Theory of Algebraic Numbers, Springer, New York, 2001, p. 503.

I add a note that related formulas can be found in

S. Sekatskii, Novel integral representations of the Riemann zeta-function and Dirichlet eta-function, close expressions.... https://arxiv.org/abs/1606.02150 (I have not checked this paper, but contains some interesting similar relations).

After several hours I find the formula in Theorem 1 of

G. T. Williams, "A new method of evaluating $\zeta(2n)$", Amer. Math. Soc, 60, (1953) 19--25.

The author of this paper think he is the first to state the result in this form.

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    $\begingroup$ it's eq. 27 of Sekatskii $\endgroup$ – Carlo Beenakker Jul 26 '18 at 18:52
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For a variation on the bilinear recursion relation in the OP that involves only sums up to $\lfloor n/2\rfloor$, rather than up to $n-1$, see [1,2].

Expanding the question to interesting and little known recursion relations for the zeta function I could mention this linear relation [3,4]:

$$(-1)^n\frac{\pi^{2n}n}{(2n+1)!}+\sum_{k=0}^{n-1}(-1)^k\frac{\pi^{2k}}{(2k+1)!}\zeta(2n-2k)=0.$$

[1] H.-T. Kuo, A recurrence formula for $\zeta(2n)$ (1949).
[2] L. Carlitz, A recurrence formula for $\zeta(2n)$ (1961).
[3] I. Song, A recursive formula for even order harmonic series (1987).
[4] M. Merca, On the Song recurrence relation for the Riemann zeta function (2016).

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