OFFSET
0,1
COMMENTS
Equals the alternating sum of a sequence of real numbers c(k) (see the Formula section). The Riemann Hypothesis is equivalent to c(k) ~ O(k^(-3/4+eps)) for all eps>0 (Báez-Duarte, 2005).
The sum without the alternating signs is Sum_{k>=0} c(k) = 1/zeta(0) = -2.
LINKS
Luis Báez-Duarte, A sequential Riesz-like criterion for the Riemann hypothesis, International Journal of Mathematics and Mathematical Sciences, Vol. 2005, No. 21 (2005), pp. 3527-3537.
Jerzy Cisło and Marek Wolf, Equivalence of Riesz and Baez-Duarte criterion for the Riemann Hypothesis, arXiv:math/0607782 [math.NT], 2006.
Jerzy Cisło and Marek Wolf, Criteria equivalent to the Riemann Hypothesis, AIP Conference Proceedings, Vol. 1079, No. 1. (2008), pp. 268-273; arXiv preprint, arXiv:0808.0640 [math.NT], 2008.
Jerzy Cisło and Marek Wolf, On the Riesz and Baez-Duarte criteria for the Riemann Hypothesis, arXiv:0807.2971 [math.NT], 2008.
Mark W. Coffey, On the coefficients of the Baez-Duarte criterion for the Riemann hypothesis and their extensions, arXiv:math-ph/0608050, 2006.
FORMULA
Equals Sum_{k>=0} (-1)^k * c(k), where c(k) = Sum_{n>=1} mu(n)*(1-1/n^2)^k/n^2 = Sum_{j=0..k} (-1)^j * binomial(k,j)/zeta(2*j+2), where mu is the Möbius function (A008683).
Equals 1 + Integral_{x>=2} (1 - 1/2^floor(x/2)) * zeta'(x)/zeta(x) dx.
EXAMPLE
0.78252798532538423457668847428378407680580097965259...
MAPLE
evalf(Sum(1/(2^k*Zeta(2*k)), k = 1..infinity), 120); # Vaclav Kotesovec, Jun 19 2021
MATHEMATICA
RealDigits[Sum[1/(2^k*Zeta[2*k]), {k, 1, 1000}], 10, 100][[1]]
PROG
(PARI) suminf(k=1, 1/(2^k*zeta(2*k)))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Amiram Eldar, Jun 18 2021
STATUS
approved