Question

Solve this...
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Answer 1

Answer:

I solved it myself.

First we note that

∑k=1∞ζ(2k+1)−1k+1=∑n=2∞∑k=1∞1(k+1)n2k+1=∑n=2∞(−1n−nlog(1−1n2))

Then

∑k=1∞ζ(2k+1)−1k+1=∑n=2∞(−1n−nlog(1−1n2))=limN→∞∑n=2N(−1n−nlog(1−1n2))=limN→∞[−HN+1−∑n=2Nnlog(n2−1)+2∑n=2Nnlog(n)]=limN→∞[−HN+1−∑n=2N(nlog(n+1)+nlog(n−1)−2nlog(n))]=limN→∞[−HN+1+log(2)−∑n=3N+1(n−1)log(n)−∑n=3N−1(n+1)log(n)+∑n=3N2nlog(n)]=limN→∞[−HN−Nlog(N+1)−(N−1)log(N)+2Nlog(N)+1+log(2)]=limN→∞(−(HN−logN)+log(2)+1−Nlog(1+1N))=limN→∞(−(HN−logN)+log(2)+O(N−1))

Since γ=limN→∞(HN−log(N)), we get

∑k=1∞ζ(2k+1)−1k+1=−γ+log(2)