Let A(p,q)=∞∑k=1(−1)k+1H(p)kkq,
where H(p)n=∑ni=1i−p, 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):
N∑n=1(−1)n−1nHn=N∑n=1(−1)n−1n2+N∑n=2(−1)n−1nHn−1=N∑n=1(−1)n−1n2+12N∑n=2n−1∑k=1(−1)n−1n(1k+1n−k)=N∑n=1(−1)n−1n2+12N∑n=2n−1∑k=1(−1)n−1k(n−k)=N∑n=1(−1)n−1n2+12N−1∑k=1N∑n=k+1(−1)n−1k(n−k)=N∑n=1(−1)n−1n2+12N−1∑k=1N−k∑n=1(−1)n+k−1kn=N∑n=1(−1)n−1n2−12N−1∑k=1(−1)k−1kN−1∑n=1(−1)n−1n+12N−1∑k=1(−1)k−1kN−1∑n=N−k+1(−1)n−1n
where, using the Alternating Series Test, we have
12|N−1∑k=1(−1)k−1kN−1∑n=N−k+1(−1)n−1n|≤12|N/2∑k=1(−1)k−1kN−1∑n=N−k+1(−1)n−1n|+12|N−1∑k=N/2(−1)k−1kN−1∑n=N−k+1(−1)n−1n|≤12⋅1⋅2N+12⋅2N⋅1=2N
Applying (2) to (1) and letting N→∞, we get
∞∑n=1(−1)n−1nHn=12ζ(2)−12log(2)2
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