Sunday, 15 September 2013

real analysis - Infinite Series sumik=1nftyleft(zeta(2)sumkn=1frac1n2right)2

Evaluate:

k=1(ζ(2)kn=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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