My Math Forum My elegant formula for the Harmonic Numbers of order k

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 November 8th, 2018, 12:16 PM #11 Newbie   Joined: May 2018 From: NYC Posts: 11 Thanks: 0 I have published my paper now: https://arxiv.org/abs/1810.07877 The manuscript is not tailored for publication, so I've seen some people/reviewers become frustrated for not understanding the argumentation, but mostly due to their own incompetence. Others have found the text to be pretty understandable though. The proof of the limits wasn't presented with rigor, but that was pointed out in the text, that's how Mathematics is done anyway (and if you disagree with me, you probably wouldn't disagree with Alon Amit). Since I didn't write the manuscript with publication in mind, I made it simple and sort of monotonous, the intent now is just to make the new results known, it's unlikely it's fit for publication as is. It's about the novelty, it's not about the presentation. The contents of this paper present formulae for the harmonic numbers of any order, for the partial sums of two Fourier series which include them, and as a consequence, a new proof for the Basel problem, as well as a proof for the closed-forms of $\displaystyle \zeta(2k)$, integer $\displaystyle k>=1$. It also has new integral representations for $\displaystyle \zeta(2k+1)$, which appear in the formulas of $\displaystyle H_{2k+1}(n)$. Assuming the closed-forms of $\displaystyle \zeta(2k)$, we are led to conclude the limits of the integrals that appear on $\displaystyle H_{2k}(n)$, and vice-versa. I am now thinking whether it's possible to extend the reasoning employed for the harmonic numbers to any harmonic progression, in principle it should be. Last edited by jrsousa2; November 8th, 2018 at 12:20 PM.
November 8th, 2018, 12:29 PM   #12
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 Originally Posted by jrsousa2 I've seen some people/reviewers become frustrated for not understanding the argumentation, but mostly due to their own incompetence.
Shinichi Mochizuki is that you?

November 8th, 2018, 12:33 PM   #13
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 Originally Posted by Maschke Shinichi Mochizuki is that you?
Definitely possible, but am not gonna say why I think so. Lol.

 November 8th, 2018, 01:00 PM #14 Newbie   Joined: May 2018 From: NYC Posts: 11 Thanks: 0 Btw, let me place the final results here as a shortcut One of the three possible formulas for the harmonic numbers is: $\displaystyle \sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\pi\int_{0} ^{1} (1-u)\cot{\pi u}\left(1-\cos{2\pi n u}\right)\,du$ But more generally, for the odd powers we get: $\displaystyle H_{2k+1}(n)=\frac{1}{2n^{2k+1}}-\frac{(-1)^{k}(2\pi)^{2k+1}}{2}\int_{0}^{1}\sum_{j=0}^{k} \frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n)\,du$ where $\displaystyle f(u,n)=\cot{\pi u}\left[1-\cos{2\pi n(1-u)}\right]$.
 November 21st, 2018, 04:42 PM #15 Newbie   Joined: May 2018 From: NYC Posts: 11 Thanks: 0 Formula for Harmonic Progression As I imagined, it's really possible to apply the same reasoning in my paper to get the sum of the Harmonic Progression. Given $a$ and $b$ integer numbers: $\sum _{k=1}^n \frac{1}{(a k+b)}=\frac{-1}{2b}+\frac{1}{2(a n+b)}+\int _0^1 2\pi u\sin{[a n\pi(1-u)]}\sin{[\pi(1-u)(a n+2b)]}\cot{[a\pi(1-u)]}du$

 Tags elegant, formula, harmonic, harmonic numbers, number theory, numbers, order, riemann, zeta

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