Saturday, October 24, 2015

sequences and series - Questions on two Formulas for zeta(s)

This question is related to the following two formulas for ζ(s).




(1) ζ(s)=1121sn=012n+1nk=0(1)k(nk)(k+1)s,s1(see ref(1) and formula (21) at ref(2))



(2) ζ(s)=1s1n=01n+1nk=0(1)k(nk)(k+1)s1(see ref(1) and formula (22) at ref(2))



Formula (1) above is claimed to converge for s1 at ref(2), but note that 1121s exhibits a complex infinity at s=1+i2πjlog(2) where jZ which seems consistent with the convergence claim at ref(1).



Question (1): Is it true that formula (1) converges for s1+i2πjlog(2) where jZ versus s1? Or is there an argument about zeros and poles cancelling each other out when formula (1) for ζ(s) is evaluated at s=1+i2πjlog(2) where jZ similar to the argument for the convergence of the right side of the functional equation \zeta(s)=2^s π^{s−1}\sin\left(\frac{π\,s}{2}\right)\,\Gamma(1−s)\,\zeta(1−s) at positive integer values of s (e.g. see Using the functional equation of the Zeta function to compute positive integer values)?







Since originally posting question (1) above, I discovered the following Wikipedia article which I believe provides some insight.



Wikipedia Article: Landau's problem with \zeta(s)=\frac{\eta(s)}{0} and solutions






Formula (2) above is claimed to be globally convergent, but seems to exhibit a significant divergence (see Figure (1) below).



Question (2): Is there an error in formula (2), or is there a conditional convergence requirement associated with formula (2) when the outer series is evaluated for a finite number of terms?




ref(1): Wikipedia Article: Riemann zeta function, Representations, Globally convergent series



ref(2): Sondow, Jonathan and Weisstein, Eric W. "Riemann Zeta Function." From MathWorld--A Wolfram Web Resource.



12/10/2018 Update:



I'm now wondering if formula (2) for \zeta(s) is perhaps only valid for s\in\mathbb{Z}.



The following plot illustrates formula (2) for \zeta(s) evaluated for the first 100 terms.







Illustration of Formula (2) for zeta(s)



Figure (1): Illustration of Formula (2) for \zeta(s)






The following discrete plot illustrates formula (2) for \zeta(s) minus \zeta(s) where formula (2) is evaluated for the first 100 terms in blue and the first 1000 terms in orange.







Discrete Plot of Formula (2) for zeta(s)



Figure (2): Discrete Plot of Formula (2) for \zeta(s) minus \zeta(s)




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