Where does this asymptote for $H_n^{(k)}$ come from?

Asked Mar 29 '20 at 06:42

Active Mar 30 '20 at 03:40

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@Claude Leibovici's answer to this Math Stack Exchange question (it's the second answer) gives an asymptote for the generalised harmonic number $H_n^{(k)}=\sum_{i=1}^n \frac{1}{i^k}$:

$$H_n^{(k)}=n^{-k} \left(-\frac{n}{k-1}+\frac{1}{2}-\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right) +\zeta (k)$$

Heuristically, this is an excellent fit. But can someone please tell me if this is a published result, and more importantly how it is derived?

edited Mar 30 '20 at 03:40

Ali Shadhar

25,498

asked Mar 29 '20 at 06:42

Richard Burke-Ward

It should follow from Euler–Maclaurin summation. – Greg Martin Mar 29 '20 at 06:51
Hi @Greg Martin - please see my updated question here: https://math.stackexchange.com/questions/3563469/inequalities-to-give-bounds-on-generalised-harmonic-numbers. Thank you. – Richard Burke-Ward Mar 30 '20 at 12:31
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@ Richard Burke-Ward Good and precise question. Here's a simple exact solution: we can easily derive this relation $H_{n}^{(k)} = \zeta(k) +\frac{(-1)^{k-1}}{(k-1)!}( \frac{d}{dn})^{k-1} H_{n}$. Then substituting the asymptotic expansion of $H_n$ immediately gives that of $H_{n}^{(k)}$. I suggest to reopen the question since IMHO the question is not a duplicate (and the solution given now here is even less so). – Dr. Wolfgang Hintze Mar 30 '20 at 16:46
Hi @Dr Wolfgang Hintze, I am have voted this for reopening. I'll look forward to accepting your answer - but I'd appreciate a little more detail. How do you arrive at that initial relation? By 'asymptotic expansion of $Hn$', do you mean McLaurin series? – Richard Burke-Ward Mar 31 '20 at 07:52
PS: I still think the other question is just a larger version of this one! And there are points available! – Richard Burke-Ward Mar 31 '20 at 07:53
@ Richard Burke-Ward In the meantime I have posted my answer here https://math.stackexchange.com/a/3602985/198592 – Dr. Wolfgang Hintze Mar 31 '20 at 09:08

Where does this asymptote for $H_n^{(k)}$ come from?

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