6

I came across the following problem in my book: $$\lim _{n \rightarrow \infty}\left(\sqrt[n+1]{\Gamma\left(\frac{1}{1}\right) \Gamma\left(\frac{1}{2}\right) \cdots \Gamma\left(\frac{1}{n+1}\right)}-\sqrt[n]{\Gamma\left(\frac{1}{1}\right) \Gamma\left(\frac{1}{2}\right) \cdots \Gamma\left(\frac{1}{n}\right)}\right)=\lim _{n \to \infty}\left(\sqrt[n+1]{\prod_{k=1}^{n+1}\Gamma\left(\frac{1}{k}\right)}-\sqrt[n]{\prod_{k=1}^{n}\Gamma\left(\frac{1}{k}\right)}\right)$$

The given answer is $ \frac{1}{e} $. However, I consistently end up with $0$ when I attempt to solve it. I am unsure where my mistake lies.

My attempt

Gauss' multiplication formula: $$\prod\limits_{k=0}^{n-1} \Gamma\left(x+\frac{k}{n}\right)=(2\pi)^{\frac{n-1}{2}}n^{\frac{1-2nz}{2}}\Gamma(nz) $$

when $z=\frac{1}{n}$ $$\lim\limits_{n\to \infty}\sqrt[n]{\prod_{k=1}^{n}\Gamma\left(\frac{1}{k}\right)}=\lim\limits_{n\to \infty}\sqrt[2n]{\frac{(2\pi)^{n-1}}{n}}=\sqrt{2\pi}$$

so $$\lim _{n \to \infty}\left(\sqrt[n+1]{\prod_{k=1}^{n+1}\Gamma\left(\frac{1}{k}\right)}-\sqrt[n]{\prod_{k=1}^{n}\Gamma\left(\frac{1}{k}\right)}\right)=0$$

I don't know how to proceed in this question since I can't use string formula as the argument of $\Gamma$ don't tend to infinity

pie
  • 8,483
  • Apologies for the bad title The latex took most space for the title I couldn't name the question "how to find ... ?" and I need to put characters in the title – pie Jan 26 '25 at 11:01
  • Setting $x=\frac 1n $ in $\prod_{k=0}^{n-1} \Gamma\left(x+\frac{k}{n}\right)$ gives $\prod_{k=1}^{n}\Gamma\left(\frac{k}{n}\right)$ and not $\prod_{k=1}^{n}\Gamma\left(\frac{1}{k}\right)$. – Martin R Jan 26 '25 at 11:53
  • 1
    @MartinR That is one of the worst silly mistakes I made :) I spent an hour trying to find the mistake but I overlooked gauss's formula and forgot to check it – pie Jan 26 '25 at 11:54
  • Remark: $a_n = \Gamma(1/n) \to \infty$ implies that $\sqrt[n]{a_1 a_2 \cdots a_n} \to \infty$, but that does not help to compute the limit of the difference. – Martin R Jan 26 '25 at 12:05

1 Answers1

5

Let $a_n=\prod_{k=1}^{n}\Gamma(1/k)$ so that $a_{n+1}/a_n=\Gamma(1/(n+1))$ and $b_n=\sqrt[n] {a_n} $. Note that $\Gamma(1/n)\to\infty$ and hence $b_n=a_n^{1/n}\to\infty$ as well and thus we need to recast $\Gamma(1/n)$ as $$\Gamma(1/n)=n\Gamma(1+1/n)$$ so that $$a_n=n! \prod_{k=1}^{n}\Gamma(1+(1/k))$$ and hence $$\frac{\sqrt[n] {a_n}} {n} =\frac{\sqrt[n] {n!}} {n} \sqrt[n] {\prod_{k=1}^{n}\Gamma(1+(1/k))}$$ and thus $b_n/n\to 1/e$.

We next have $$\frac{b_{n+1}}{b_n}=\frac{b_{n+1}}{n+1}\frac{n+1}{n}\frac{n}{b_n}\to 1 $$ and thus $$b_{n+1}-b_n=b_n\left(\frac{b_{n+1}}{b_n}-1\right)=b_n\dfrac{(b_{n+1}/b_n)-1} {\log(b_{n+1}/b_n)}\cdot\log\frac{b_{n+1}}{b_n} $$ and the desired limit is equal to the limit of $$b_n\log\frac{b_{n+1}}{b_n}=\frac{b_n}{n}\cdot n\log \frac{b_{n+1}}{b_n}$$ and our job is done (ie required limit is established as $1/e$) if we show that $$n\log \frac{b_{n+1}}{b_n}\to 1$$ But this expression equals $$\log\frac{a_{n+1}}{a_nb_{n+1}}=\log\frac{n+1}{b_{n+1}}+\log\Gamma\left(\frac {n+2}{n+1}\right)$$ (check this) and the above tends to $$\log e+\log 1=1$$ and we are done.

The above is a special case of a more general theorem discussed here.

Paramanand Singh
  • 92,128