Question

5Cbigg%28%20%5Cfrac%7B1%7D%7Bk%7D%20%20%5Cbigg%29%5CGamma%20%5Cbigg%28%20%20%5Cfrac%7B2%7D%7Bk%7D%20%5Cbigg%29%5CGamma%20%5Cbigg%28%20%20%5Cfrac%7B3%7D%7Bk%7D%20%5Cbigg%29%20%5Cdots%5CGamma%20%5Cbigg%28%20%5Cfrac%7Bk%7D%7Bk%7D%20%20%5Cbigg%29%7D%20%20%5C%5C%20" id="TexFormula1" title=" \rm \lim_{k \to \infty } \sqrt[ k]{ \Gamma \bigg( \frac{1}{k} \bigg)\Gamma \bigg( \frac{2}{k} \bigg)\Gamma \bigg( \frac{3}{k} \bigg) \dots\Gamma \bigg( \frac{k}{k} \bigg)} \\ " alt=" \rm \lim_{k \to \infty } \sqrt[ k]{ \Gamma \bigg( \frac{1}{k} \bigg)\Gamma \bigg( \frac{2}{k} \bigg)\Gamma \bigg( \frac{3}{k} \bigg) \dots\Gamma \bigg( \frac{k}{k} \bigg)} \\ " align="absmiddle" class="latex-formula">​

Answer 1

We have

$\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)$

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$