I want to show that we can continue Riemann's zeta function to Re(s)>0, s≠1 by the following formula
(1−21−s)ζ(s)=(1−212s)(11s+12s+…)=11s+12s+…−2(12s+14s+…)=11s−12s+13s−14s+…=∞∑n=1(−1)n−11ns.
In order to do that, I need to show that the series converges for Re(s)>0, except s=2kπiln2+1, k∈Z, which are removable singularities. Any ideas on how to do that?
Answer
Write the Taylor expansion of (1+x)a around x=0, and n−s−(n+1)−s=n−s(1−(n+1n)−s)=n−s(1−(1+1n)−s)=O(n−s−1)
which is summable for Re(s)>0, hence by grouping the terms by two :
η(s)=(1−21−s)ζ(s)=∞∑n=1(2n−1)−s−(2n)−s is absolutely convergent for any Re(s)>0. (η(s) is called the Dirichlet η function )Other way, integrate by part (with δ(x) the Dirac delta distribution)∞∑n=1ann−s=∫∞1−ϵ(∞∑n=1anδ(x−n))x−sdx=s∫∞1(∑n≤xan)x−s−1dx here an=(−1)n+1 hence ∑n≤xan=1 or 0 and η(s)=s∫∞1(∑n≤x(−1)n+1)x−s−1dx converges for Re(s)>0 (it's called the Perron's formula )
Last way, use that n−sΓ(s)=n−s∫∞0xs−1e−xdx=∫∞0ys−1e−nydy (change of variable y=nx, and Γ(s) the Gamma function) hence :
Γ(s)∞∑n=1ann−s=∫∞0ys−1(∞∑n=1ane−ny)dy
(by using the absolute/dominated convergence theorem for exchanging ∑ and ∫)here ∑∞n=1(−1)n+1e−ny=1ey+1 hence
η(s)Γ(s)=∫∞0ys−1ey+1dy
which has no singularity for Re(s)>0.Last last way : η(ϵ)=∞∑n=1(−1)n+1n−ϵ
is a convergent alternated series for ϵ>0, hence by the abscissa of convergence theorem for Dirichlet series, we get that ∑∞n=1(−1)n+1n−s converges for any Re(s)>ϵ, and since ϵ is arbitrary small, for any Re(s)>0.
(if someone knows another way...)
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