Tuesday, 12 April 2016

real analysis - Proving an alternating Euler sum: suminftyk=1frac(1)k+1Hkk=frac12zeta(2)frac12log22



Let A(p,q)=k=1(1)k+1H(p)kkq,


where H(p)n=ni=1ip, the nth p-harmonic number. The A(p,q)'s are known as alternating Euler sums.





Can someone provide a nice proof that
A(1,1)=k=1(1)k+1Hkk=12ζ(2)12log22?




I worked for a while on this today but was unsuccessful. Summation by parts, swapping the order of summation, and approximating Hk by logk were my best ideas, but I could not get any of them to work. (Perhaps someone else can?) I would like a nice proof in order to complete my answer here.



Bonus points for proving A(1,2)=58ζ(3) and A(2,1)=ζ(3)12ζ(2)log2, as those are the other two alternating Euler sums needed to complete my answer.






Added: I'm going to change the accepted answer to robjohn's A(1,1) calculation as a proxy for the three answers he gave here. Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying. I am pleased to see that it can be used to do the A(1,1), A(1,2), and A(2,1) derivations.

Answer



A(1,1):
Nn=1(1)n1nHn=Nn=1(1)n1n2+Nn=2(1)n1nHn1=Nn=1(1)n1n2+12Nn=2n1k=1(1)n1n(1k+1nk)=Nn=1(1)n1n2+12Nn=2n1k=1(1)n1k(nk)=Nn=1(1)n1n2+12N1k=1Nn=k+1(1)n1k(nk)=Nn=1(1)n1n2+12N1k=1Nkn=1(1)n+k1kn=Nn=1(1)n1n212N1k=1(1)k1kN1n=1(1)n1n+12N1k=1(1)k1kN1n=Nk+1(1)n1n


where, using the Alternating Series Test, we have
12|N1k=1(1)k1kN1n=Nk+1(1)n1n|12|N/2k=1(1)k1kN1n=Nk+1(1)n1n|+12|N1k=N/2(1)k1kN1n=Nk+1(1)n1n|1212N+122N1=2N

Applying (2) to (1) and letting N, we get
n=1(1)n1nHn=12ζ(2)12log(2)2



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