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Assume that $f$ is a function such that $\lim_{\varepsilon \rightarrow 0^+} f(\varepsilon) = 1$. Is there an $f$ such that

$$\lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{1}{n} f(n \varepsilon)$$

exists? In another question, I found an $f$ such that

$$\lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(n \varepsilon) = \zeta(-s)$$

for $s \neq -1$, namely

$$f(x) = \mathrm{e}^{-x}\left(1-\frac{x}{s+1}\right)$$

This regulator was obtained by seeking an $f$ such that

$$\int_0^\infty x^s f(x) \,\mathrm{d}x$$

for $s > -1$. Hence we can similarly ask whether there is an $f$ such that

$$\int_0^\infty \frac{1}{x} f(x) \,\mathrm{d}x = 0$$

or whether no such $f$ exists. Here I describe a way of regularizing the harmonic series that is based on the above regulator, but does not take the same form.

user76284
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1 Answers1

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I don't have the reputation to comment, so I'll mark my answer as community wiki since it's not a full answer:

I'm not aware of a regulator of exactly the type you are looking for, but the harmonic series is logarithmically divergent, and can be regulated by a kind of "minimal subtraction" scheme to obtain the Euler-Mascheroni constant as a finite answer.

It may be possible to modify this to form a function of the type you want.

  • Thank you for your reply. Indeed I’m expecting such a regulator, if it exists, to yield $\gamma$ as the result. I’ll keep searching for a function of the right form. – user76284 Jun 28 '19 at 01:22
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    @user76284 Consider any such $f$. Take $g(x)=f(x)+\frac{2\alpha}\pi\sin(x)$. Then we have $\lim_{\varepsilon\to0^+}g(\varepsilon)=1$ and $\lim_{\varepsilon\to0^+}\sum_{n=1}^\infty\frac1ng(n\varepsilon)=\alpha+\lim_{\varepsilon\to0^+}\sum_{n=1}^\infty\frac1nf(n\varepsilon)$. – Simply Beautiful Art Jun 28 '19 at 01:33
  • @SimplyBeautifulArt On further thought, I think there's something fishy about that regularization. Naively, one would think it's okay because $\frac{2\alpha}{\pi} \sin n \varepsilon \rightarrow 0$ as $\varepsilon \rightarrow 0^+$. But, by the same token, $n \frac{2\alpha}{\pi} \sin n \varepsilon \rightarrow 0$ as $\varepsilon \rightarrow 0^+$. – user76284 Jul 21 '19 at 08:59
  • @SimplyBeautifulArt Therefore, the same reasoning would allow us to "regularize" the already-convergent series $\sum_{n=1}^\infty \frac{1}{n^2}$ as $\lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty \frac{1}{n^2}(1 + n \frac{2\alpha}{\pi} \sin n \varepsilon) = \frac{\pi^2}{6} + \alpha$, which is the wrong answer (for $\alpha \neq 0$). – user76284 Jul 21 '19 at 08:59
  • @user76284 that's the point. – Simply Beautiful Art Jul 22 '19 at 09:48
  • Well in your example you don't have $f(n\epsilon)$ though... – Simply Beautiful Art Jul 22 '19 at 10:24
  • @SimplyBeautifulArt Something else I noticed: $\sum_{n=1}^\infty \frac{1}{n} g(n \varepsilon)$ does not converge uniformly to $1 - \frac{\varepsilon}{\pi}$ on $\varepsilon > 0$. I think that has something to do with the additional term being "illegitimate" in some sense (that I'm trying to work out). – user76284 Apr 14 '21 at 20:16