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

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \(q\)

Log 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

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.

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply