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I have a question about limits that are divergent. In some case we can give them results like in case of $\lim\limits_{n \rightarrow \infty} \Gamma (n)=\sqrt {2\pi} $. It can be goten by use of zeta function as follow

$\displaystyle \lim\limits_{n \rightarrow \infty} \ln (\Gamma (n))=\sum_{n=1}^{\infty}ln (n)=-\lim\limits_{s \rightarrow 0} \frac {d}{ds} \zeta (s)=ln (\sqrt {2\pi})$

Is it possible to give such anwser to all limits? Is there some theorem that generalise such concept?

Wreior
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  • What does $\Gamma$ represent? Would you also please add more detail? – Math-fun Apr 02 '23 at 18:54
  • Is that $\Gamma$ the Gamma function? If so, I don't know of an interpretation where that limit exists. – Randall Apr 02 '23 at 18:55
  • Yes, it's standard gamma function – Wreior Apr 02 '23 at 18:56
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    Can you provide a reference for a definition for $\lim \Gamma (n)$ under which the limit above holds? Also - there’s a nice way to generalize limits using a linear functional. It’s called “Banach limit “, but as far as I know it’s only defined for bounded sequences. – NG_ Apr 02 '23 at 19:06
  • Besides what @NG_ said, another way of assigning a limit (in all cases, where the range is real-valued or complex-valued) is to assign to the sequence the set of all subsequence limits (in the extended real line if real-valued; in the Riemann sphere if complex-valued). In this case the "single mathematical object" that is the "limit" is a certain set of numbers. The sets in question are always closed sets, so an arbitrary set of numbers cannot be such a "limit" (see 1 and 2). – Dave L. Renfro Apr 02 '23 at 19:34
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    @Wreior. I assume you are interested in zeta-regularized values of limits, sums, and products. See "The the zeta-regularized product over all primes is $4\pi^2$" for example. – bob Apr 02 '23 at 22:40

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Here's one way to interpret the question-- we want to find the constant term in the expansion of a certain function $f(z)$ at $z = \pm \infty$.

One way we can do this is to write $f(z)$ as a power series
$$f(z) = \sum_{n=1}^\infty a_n (-x)^n$$ Then, subject to growth conditions, we can write $f(z)$ as the contour integral $$f(z) = \int_C \frac{\csc(\pi n)}{2i} a_n x^n dn$$ Switching the direction of this contour and assuming that $a_n$ has no poles gives the sum $$- \sum_{n=0}^\infty a_{-n} (-x)^{-n}$$ Which, as $x \to \infty$ is just $-a_0$. Therefore, we just need to study the pole at $a_0$. Let's consider the example of $\ln \Gamma$. $\ln \Gamma$ has a well known Taylor series of $a_n = \frac{\zeta(k)}{k}$. We have $2 \pi i \text{Res}_{z=0}\left( \frac{\zeta(z)}{z} \frac{\csc(\pi z)}{2 i}\right) = -\frac{\ln(2 \pi)}{2} - \frac{\ln(x)}{2}$. The negative of the constant part of this is $ \frac{\ln(2 \pi)}{2}$. This same result can be obtained by directly looking at the expansion of the log gamma function at $\infty$ $$\ln \Gamma (z) = z\ln z - z - \frac{1}{2} \ln (z) + \frac{1}{2} \ln(2 \pi) + O\left(\frac{1}{z}\right)$$

If we instead looked at the sum $$-\gamma x + \sum_{n=2}^\infty \zeta(k) (-x)^k$$ We would obtain that $a_0 = -\zeta(0) = \frac{1}{2}$. This result is alternatively easily obtained by looking at the constant term in the expansion of $-x\frac{d}{dx} \ln \Gamma (x)$.

There are a number of results that this method can obtain with this method. For instance, it gives us that the constant term $$\sum_{n=1}^\infty \frac{1}{n^2} \frac{(-x)^n }{n!} \cong -\frac{1}{2}\left(\gamma^2+ \zeta(2)\right)$$ More specifically, it tells us that $$\lim_{x \to \infty} \sum_{n=1}^\infty \frac{1}{n^2} \frac{(-x)^n }{n!} + \gamma \ln(x) + \frac{1}{2} \ln(x)^2 = -\frac{1}{2}\left(\gamma^2+ \zeta(2)\right)$$ Alternatively, we have the sum $$\sum_{n=2}^\infty \frac{\zeta(n)}{n^2}(-x)^n \cong \frac{5 \pi^2}{48} + \ln(\sqrt{2 \pi})^2-\frac{\gamma^2}{4} - \frac{\gamma_1}{2} $$ And by looking closely at the residues we could find exactly the needed terms to subtract to write that congruence between the constant terms as a limit.

Caleb Briggs
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