Proving that π332=1−∑k=1∞2k(2k+1)ζ(2k+2)42k+2

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Karly Logan · Accepted Answer

Yes, we can prove it. We can change the order of summation in  ∑k=1∞2k(2k+1)ζ(2k+2)42k+2=∑k=1∞2k(2k+1)42k+2∑n=1∞1n2k+2  ∑n=1∞∑k=1∞2k(2k+1)(4n)2k+2  ∑n=1∞r(4n)  where, for |z|&amp;gt;1, we define  r(z)=∑k=1∞1z2k=1z2−1=12(1z−1−1z+1)  Differentiating yields r(z)=1(z−1)3−1(z+1)3, so  1−∑k=1∞2k(2k+1)ζ(2k+2)4k+2=∑v=0∞(−1)v(2v+1)3  Using the partial fraction decomposition of 1cos⁡z:  ∑v=0∞(−1)v(2v+1)3=−π332E2=π332

nebajcioz · Accepted Answer

Starting from the Laurent series of the cotangent function:  πzcot⁡(πz)=1−2∑k=0∞ζ(2k+2)z2k+2  apply the differential operator:  D^=z2d2dz2−2zddz+2  to get:  z3π3cot⁡(πz)(1+cot⁡(πz)2)=1−∑k=0∞2k(2k+1)ζ(2k+2)z2k+2  which, by the ratio test, has a radius of convergence of |z|&amp;lt;1. Then from:  z=14, cot⁡(π4)=1  we have:  π332=1−∑k=0∞2k(2k+1)ζ(2k+2)42k+2

RizerMix · Accepted Answer

S=∑k=1∞2k(2k+1)ζ(2k+2)42k+2=∑k=1∞(2k+1)!ζ(2k+2)(2k−1)!42k+2 =∑k=1∞(1/4)2k+2(2k−1)!∫0∞x2k+1ex−1dx=4−3∫0∞x2ex−1(∑k=1∞(x/4)2k−1(2k−1)!)dx =4−3∫0∞x2ex−1sinh⁡(x4)dx=4−32∫0∞x21−e−x(e−34x−e−54x)dx =4−3[ζ(3,34)−ζ(3,54)]=1−4−3[ζ(3,14−ζ(3,34)] =1−β(3)=1−π332

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

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