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

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.

## 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 \(q\) – Aditya Sharma · 3 months ago

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(q-2) (x) Li(2) (x))/x dx – Ishan Singh · 3 months agoLog in to reply

here, in a similar way before. – Ishan Singh · 3 months ago

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 SumLog in to reply

– Aditya Sharma · 3 months ago

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

– Ishan Singh · 3 months ago

There is another mistake in your formula, from what you have done, there should be a \((-1)^k\) inside the summationLog in to reply

– Aditya Sharma · 3 months ago

The formula for the alternate harmonic sum is fine .it's written (-x) and not xLog in to reply

– Ishan Singh · 3 months ago

As I said, there is a sign error in the second red line, it should be \(+ \int_{0}^{1} \dfrac{\operatorname{Li}_{q-2}(-x) \operatorname{Li}_2(-x)}{x} \mathrm{d}x\) and not \(- \int_{0}^{1} \dfrac{\operatorname{Li}_{q-2}(-x) \operatorname{Li}_2(-x)}{x} \mathrm{d}x\) (even if we take (-x)).Log in to reply

– Aditya Sharma · 3 months ago

Thanks for that , tou said there should be a \((-1)^k\) in your summation. Summation reffers only to the integral representation of the alternate harmonic number. But that's fine. I said there is a \((-x)\) in that formulaeLog in to reply

here. – Ishan Singh · 3 months ago

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

– Ishan Singh · 3 months ago

No, it is for all \(q\), even or odd.Log in to reply