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Finding Euler

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

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.

Note by Ishan Singh
9 months, 1 week ago

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

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 Narayan Sharma · 9 months ago

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@Aditya Narayan Sharma 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 dx Ishan Singh · 9 months ago

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@Aditya Narayan Sharma 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. Ishan Singh · 9 months ago

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@Ishan Singh So I may assume that the result as quoted by you of this summation is for the even \(q\) ? Aditya Narayan Sharma · 9 months ago

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@Aditya Narayan Sharma There is another mistake in your formula, from what you have done, there should be a \((-1)^k\) inside the summation Ishan Singh · 9 months ago

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@Ishan Singh The formula for the alternate harmonic sum is fine .it's written (-x) and not x Aditya Narayan Sharma · 9 months ago

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@Aditya Narayan Sharma 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)). Ishan Singh · 9 months ago

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@Ishan Singh 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 formulae Aditya Narayan Sharma · 9 months ago

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@Aditya Narayan Sharma I meant in the eta summation, there should be a \((-1)^k\) in the sum. Compare with original Euler Sum in my answer here. Ishan Singh · 9 months ago

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@Aditya Narayan Sharma No, it is for all \(q\), even or odd. Ishan Singh · 9 months ago

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