みさき氏の級数を示します！
∑n=1∞Hn(2)n4n(2nn)=32ζ(3)
以下、Cr=2−2r(2rr)=(2r−1)!!(2r)!!とする。
∑n=1∞Hn(2)n4n(2nn)=∑n=1∞Cnn∑m=1n1m2=∑0<m≤nCnm2n=∑0<m≤n∫01tm−1dtCnmn=∫01dtt∑0<m≤ntmCnmn=∫01dtt∑0<m≤n∫0txm−1dxCnn=∫01dtt∫0tdx∑0≤m<nxmCnn=∫01dtt∫0tdx∑0<n1−xn1−xCnn=∫01dtt∫0tdx1−x∑0<n(1−xn)Cnn=∫01dtt∫0tdx1−x∫x1(1−u)−1/2−1udu=∫0<x<t<1,0<x<u<1dttdx1−x(1−u)−1/2−1udu=∫0<x<t<u<1dttdx1−x(1−u)−1/2−1udu+∫0<x<u<t<1dttdx1−x(1−u)−1/2−1udu=∫01(1−u)−1/2−1udu∫0udtt∫0tdx1−x+∫01dtt∫0t(1−u)−1/2−1udu∫0udx1−x=−4∫01du−1−u∫u1(11+t+1−1−t)dt∫t1dxx−4∫01(11+t+1−1−t)dt∫t1du−1−u∫u1dxx=−4∫01a(b0+b1)b1+ab1(b0+b1)=−4∫01z2,0z1,1+z2,1z1,1+z2,1z1,0+z2,1z1,1=−4(ζ(1―,2)+ζ(1―,2―)+2ζ(1,2―))=−4(ζ(3)−π2ln24+π2ln24−138ζ(3)+14ζ(3))=32ζ(3)
式の分量が多かった…
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