Evaluate:
∞∑k=1(ζ(2)−k∑n=11n2)2
Recognizing that ζ(2)−∑kn=11n2 can be written as ψ1(1+k) where ψ1(z) is the trigamma function,
What remains to be done is to evaluate:
∞∑k=1ψ21(k+1)
Mathematica could not evaluate it in a closed form but the source assures that it exists.
If you liked this problem check out Hard Definite integral involving the Zeta function.
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