\rm \lim_(k \to \infty ) \sqrt[ k]{ \Gamma \bigg( (1)/(k) \bigg)\Gamma \bigg( (2)/(k) \bigg)\Gamma \bigg( (3)/(k) \bigg) \dots\Gamma \bigg( (k)/(k) \bigg)} \\​

Question

`\rm \lim_(k \to \infty ) \sqrt[ k]{ \Gamma \bigg( (1)/(k) \bigg)\Gamma \bigg( (2)/(k) \bigg)\Gamma \bigg( (3)/(k) \bigg) \dots\Gamma \bigg( (k)/(k) \bigg)} \\`​

Answer 1

We have

`\sqrt[k]{\Gamma\left(\frac1k\right) \Gamma\left(\frac2k\right) \cdots \Gamma\left(\frac kk\right)} \\\\ = \exp\left(\frac{\ln\left(\Gamma\left(\frac1k\right) \Gamma\left(\frac2k\right) \cdots \Gamma\left(\frac kk\right)\right)}k\right) \\\\ = \exp\left(\frac{\ln\left(\Gamma\left(\frac1k\right)\right)+\ln\left( \Gamma\left(\frac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\frac 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(\frac1k\right) \Gamma\left(\frac2k\right) \cdots \Gamma\left(\frac 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(\frac{\ln(2\pi)}2}\right) = \boxed{√(2\pi)}`