Proving that \frac{\pi^3}{32}=1-\sum_{k=1}^\infty\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}

spiderifilms6e 2022-01-22 Answered
Proving that π332=1k=12k(2k+1)ζ(2k+2)42k+2
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Karly Logan
Answered 2022-01-23 Author has 11 answers
Yes, we can prove it. We can change the order of summation in
k=12k(2k+1)ζ(2k+2)42k+2=k=12k(2k+1)42k+2n=11n2k+2
n=1k=12k(2k+1)(4n)2k+2
n=1r(4n)
where, for |z|>1, we define
r(z)=k=11z2k=1z21=12(1z11z+1)
Differentiating yields r(z)=1(z1)31(z+1)3, so
1k=12k(2k+1)ζ(2k+2)4k+2=v=0(1)v(2v+1)3
Using the partial fraction decomposition of 1cosz:
v=0(1)v(2v+1)3=π332E2=π332
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nebajcioz
Answered 2022-01-24 Author has 15 answers
Starting from the Laurent series of the cotangent function:
πzcot(πz)=12k=0ζ(2k+2)z2k+2
apply the differential operator:
D^=z2d2dz22zddz+2
to get:
z3π3cot(πz)(1+cot(πz)2)=1k=02k(2k+1)ζ(2k+2)z2k+2
which, by the ratio test, has a radius of convergence of |z|<1. Then from:
z=14, cot(π4)=1
we have:
π332=1k=02k(2k+1)ζ(2k+2)42k+2
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RizerMix
Answered 2022-01-27 Author has 438 answers
S=k=12k(2k+1)ζ(2k+2)42k+2=k=1(2k+1)!ζ(2k+2)(2k1)!42k+2 =k=1(1/4)2k+2(2k1)!0x2k+1ex1dx=430x2ex1(k=1(x/4)2k1(2k1)!)dx =430x2ex1sinh(x4)dx=4320x21ex(e34xe54x)dx =43[ζ(3,34)ζ(3,54)]=143[ζ(3,14ζ(3,34)] =1β(3)=1π332
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