After numerical analysis it seems that
π332=1−∞∑k=12k(2k+1)ζ(2k+2)42k+2
Could someone prove the validity of such identity?
Answer
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+2∞∑n=11n2k+2=∞∑n=1∞∑k=12k(2k+1)(4n)2k+2=∞∑n=1r″(4n),
where, for |z|>1, we define
r(z)=∞∑k=11z2k=1z2−1=12(1z−1−1z+1).
Differentiating yields r″(z)=1(z−1)3−1(z+1)3, so
1−∞∑k=12k(2k+1)ζ(2k+2)42k+2=∞∑ν=0(−1)ν(2ν+1)3,
and the latter sum is by an earlier answer using the partial fraction decomposition of 1cosz:
∞∑ν=0(−1)ν(2ν+1)3=−π332E2=π332.
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