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Specifically,

$$H_m^{(2n)} \approx\ ?$$

and

$$H_m^{(4n)} \approx\ ?$$

where $(m, n)$ $\in \mathbb N_{>1}$

I would not like to use special functions like the (Riemann zeta function) unless they themselves have polynomial expressions approximating them with shrinking error terms (see https://math.stackexchange.com/a/1583465/322359)

You may wish to reference : http://mathworld.wolfram.com/HarmonicNumber.html

This is motivated by my work involving summation of generalized harmonic numbers and series/alternate representations there in. Summation is not too difficult, series/alternate representations are a bit rough for me.

Take a look at work here: https://math.stackexchange.com/a/1583465/322359 which seems to solve the issue entirely using the Euler-Maclaurin formula, only we have that pesky (Riemann) zeta function as one of the terms...

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  • $\begingroup$ Take the asymptotic form. For instance for $m \to \infty$ we have $H_{m}^{(2n)} \simeq \zeta (2 n)+\left(\frac{m}{1-2 n}-\frac{n}{6 m}+\frac{1}{2}\right) m^{-2 n}$ But I see no way to avoid zeta functions (and I don't see why they should be avoided). $\endgroup$ – Dr. Wolfgang Hintze May 11 at 16:36
  • $\begingroup$ @Dr.WolfgangHintze We can use the zeta functions, I just don't know of any polynomial expansions for them. I'm looking for something polynomial, with decreasing error term ideally. I suspect a polynomial expansion for $\zeta(2n)$ would be quite well known using Euler-Maclaurin or something. $\endgroup$ – user3108815 May 11 at 23:25
  • $\begingroup$ You can write $\zeta(2n) = 1 + 1/2^(2n) + 1/3^(2n)+...$ which is a fairly good approximation for $n \ge 1$. So what is left open of your question? $\endgroup$ – Dr. Wolfgang Hintze May 12 at 7:33
  • $\begingroup$ @Dr.WolfgangHintze You have not approximated it, you have just restated one of the definitions. You can truncate the definition's infinite series to get an approximation is what you're saying, but that's hardly polynomial, seems quite linear to me. If I can combine all the terms involving $n$ because they all have the same power, it is not really a polynomial. $\endgroup$ – user3108815 May 13 at 21:37

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