\[ \sum_{n=1}^{\infty} \dfrac{\overline{H_{n}}}{n^q} = \zeta(q) \log 2 - \dfrac{q}{2} \zeta(q+1) + \sum_{k=1}^{q} \eta(k) \eta(q-k+1) \]

Prove the equation above.

**Notation :** \(\displaystyle \overline{H_{n}} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\) denotes the Alternating Harmonic Number.

\(\displaystyle \zeta(s) = \sum_{k=1}^{\infty} \dfrac{1}{k^s} \) denotes the Riemann Zeta Function.

\(\displaystyle \eta(s) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^s} \) denotes the Dirchlet Eta Function.

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## Comments

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TopNewestSince \(\displaystyle \overline{H_n} = \sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}=\int_{0}^{1} \frac{1-(-x)^n}{1+x}dx\) ,

Summing both sides we obtain \(\displaystyle \sum_{n=1}^{\infty} \frac{\overline{H_n}}{n^q} = \int_{0}^{1}\frac{\zeta(q)}{1+x}dx - \int_{0}^{1}\frac{Li_q(-x)}{1+x}dx=\zeta(q)\ln 2 -\underbrace{\int_{0}^{1}\frac{Li_q(-x)}{1+x}dx}_{\color{red}{I}}\)

Now applying IBP we have, \(\displaystyle \color{red}{I} = \int_{0}^{1}\frac{Li_q(-x)}{1+x}dx = [Li_q(-x)\ln(1+x)]_{0}^{1}+\int_{0}^{1} \frac{Li_{q-1}(-x)\ln(1+x)}{x}dx\)

Since \(\displaystyle Li_q(-1)=-\eta(q),\eta(1)=\ln 2\) and applying IBP again,

\(\displaystyle \color{red}{I} = -\eta(q)\eta(1)-[Li_{q-1}(-x)Li_2(-x)]_{0}^{1} - \int_{0}^{1} \frac{Li_{q-2}(-x)Li_2(-x)}{x}dx = -\eta(q)\eta(1)-\eta(q-1)\eta(2)+ \int_{0}^{1} \frac{Li_{q-2}(-x)Li_2(-x)}{x}dx\)

Similarly applying IBP \(p\) times , we will get

\(\displaystyle \color{red}{I}=-\sum_{r=1}^{q}\eta(q-r+1)\eta(r) -\eta(q+1)\)

Therefore , \(\displaystyle \sum_{n=1}^{\infty} \frac{\overline{H_n}}{n^q}=\zeta(q)\ln2+\sum_{k=1}^{q}\eta(q-k+1)\eta(k)+\eta(q+1)\) for odd \(q\)

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Note that when you apply IBP, every time the sign of the integral changes. So your formula will regenerate the integral only if \(q\) is odd. In other words, your formula is correct, when \(q\) is odd. For instance, I have solved the regular Euler Sum here, in a similar way before.

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So I may assume that the result as quoted by you of this summation is for the even \(q\) ?

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here.

I meant in the eta summation, there should be a \((-1)^k\) in the sum. Compare with original Euler Sum in my answerLog in to reply

Also, there seems to be a sign error in the second red line, it should be + integral 0 to 1 ((Li)

(q-2) (x) Li(2) (x))/x dxLog in to reply