βn:=(2nn)22n,A:=∑n=0∞βn3,B:=∑n=1∞1n3βn3,C:=∑n=0∞(−1)n(n+12)2とする.
∑n=0∞βn3=A∑n=0∞βn3∑k=1n1k=2A(π3−ln2)∑n=0∞βn3∑k=1n1k2=A(C−π23)∑n=0∞βn3∑k=1n1k3=?∑n=0∞βn3∑k=1n1k2βk=Aπ22−Bπ∑n=0∞βn3∑k=1n1k3βk=?∑n=0∞βn3∑k=1n1k2βk2=(π2−2C)A∑n=0∞βn3∑k=1n1k3βk2=(π3−14ζ(3))A−B∑n=0∞βn3∑k=1n1k3βk3=A(B−Aπ33)
∑n=0∞βn3∑k=0n−11k+12=Aπ3∑n=0∞βn3∑k=0n−11(k+12)2=(π22−C)A∑n=0∞βn3∑k=0n−11(k+12)3=?∑n=0∞βn3∑k=0n−11(k+12)2βk=(2C−π2)A+Bπ∑n=0∞βn3∑k=0n−11(k+12)3βk=?∑n=0∞βn3∑k=0n−11(k+12)2βk2=(π2−2C)A∑n=0∞βn3∑k=0n−11(k+12)3βk2=(28ζ(3)−π3)A∑n=0∞βn3∑k=0n−11(k+12)3βk3=A(B−Aπ33)
∑n=1∞1n2∑k=0n−1(k+14)βk3=A−1A(12+Cπ2)∑n=1∞1n3∑k=0n−1(k+14)βk3=?∑n=1∞(1n2βn∑k=0n−1(k+14)βk3−2πn)=?∑n=1∞1n3βn∑k=0n−1(k+14)βk3=?∑n=1∞1n3βn2∑k=0n−1(k+14)βk3=?∑n=1∞(1n3βn3∑k=0n−1(k+14)βk3−2n)=4ln2−π∑n=0∞1(n+12)2∑k=0n(k+14)βk3=Cπ2A+A∑n=0∞1(n+12)3∑k=0n(k+14)βk3=?∑n=0∞(1(n+12)2βn∑k=0n(k+14)βk3−2π(n+1))=?∑n=0∞1(n+12)3βn∑k=0n(k+14)βk3=?∑n=0∞1(n+12)3βn2∑k=0n(k+14)βk3=1A(2Cπ+π)∑n=0∞(1(n+12)3βn3∑k=0n(k+14)βk3−2n+1)=4ln2−π
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