Here http://integralsandseries.prophpbb.com/topic119.html
We came across the following harmonic sum
∑k≥1(−1)k−1k2H(2)k
Note that we define
H(2)k=k∑n≥11n2
Also we have
ψ1(k+1)=ζ(2)−H(2)k
Any ideas how to evaluate (1) ?
Answer
A related problem. You can have the following identity
∞∑k=1(−1)k−1H(2)kk2=3716ζ(4)+2∞∑k=1(−1)kHkk3∼0.7843781621.
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