Prove that: \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^mE_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}

spiderifilms6e 2022-01-21 Answered
Prove that:
n=0(1)n(2n+1)2m+1=(1)mE2mπ2m+14m+1(2m)!
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Answers (3)

nick1337
Answered 2022-01-27 Author has 575 answers
star233
Answered 2022-01-27 Author has 208 answers
The Dirichlet beta function is defined as β(2m+1)=n=0(1)n(2n+1)2m+1 Then β(1)=π4 and β(2m+1)=k=1m(π2/4)k(2k)!β(2m2k+1) (1) If we reindex recursion derived below, we get that the even Euler numbers are defined by E0=1 and E2m=k=1m((2m),(2k))E2m2k (2) then notice that (1) is the same as (2) if we set β(2m+1)=(1)mE2mπ2m+14m+1(2m)!
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RizerMix
Answered 2022-01-27 Author has 438 answers
My proof works through the following lines: the LHS is: 1(2m)!01)logx)2m1+x2dx=12(2m)!0+(logx)2m1+x2dx so we just need to compute: d2mdk2m0+xk1+x2|k=0 but: 0+x1/r1+x2dx=r0+yr1+y2rdy =π/2cos(π/(2r)) by the residue theorem, so n=0+(1)n(2n+1)2m+1=E2m2(2m)!(π2)2m+1 where E2m is just the absolute value of an Euler number, that belongs to N.
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