The limits of the gamma function

Question

Let $$g(m)=\frac{\left[\Gamma(1+\frac{1}{m})\right]^2}{\Gamma(1+\frac{2}{m})},$$ how to consider the limit of $g(m)$ when $m\rightarrow 0^+$ ?If the limit exsits ,calculate it.

Answer 1

Using $\displaystyle \Gamma(1+x)=\lim\limits_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n \left(1+\frac{x}{k}\right)}$ one gets $\displaystyle \frac{\Gamma(1+x)^2}{\Gamma(1+2x)}=\prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k+x}\right)^2\right)$ .

Then $x\to\infty$ .

You can also use $\displaystyle 0<\frac{n!^2}{(2n)!}\leq\frac{1}{2^n}$ for $n\in\mathbb{N}$ which is easier to proof (e.g. induction).