Lower bound for $\left(\frac{\Gamma(a/2)}{\Gamma(a)}\right)^{1/a}$

Question

Let $a\in\mathbb N, a\ge 2$. During calculus class, the lecturer said that $$\left(\frac{\Gamma(a/2)}{\Gamma(a)}\right)^{1/a}\ge \frac{1}{2a},$$ where $\Gamma$ denotes the Gamma-function: https://en.wikipedia.org/wiki/Gamma_function.

When I asked for a motivation, he simply replied that it easily comes from the properties of the Gamma function.

I am trying to use the property $\Gamma(a+1)= a\Gamma(a)$ trying to write that quotient in a smart way, but with any success so far.

Anyone could please help me with that? Thank you in advance.

Answer 1

This is true but you can have much better bounds.

Take logarithms, use Stirling approximation to have $$\log \left(\left(\frac{\Gamma \left(\frac{a}{2}\right)}{\Gamma (a)}\right)^{\frac{1}{a}}\right)=\frac{1}{2} (-\log (a)+1-\log (2))+\frac{\log (2)}{2 a}+O\left(\frac{1}{a^2}\right)$$

Exponentiate $$\left(\frac{\Gamma \left(\frac{a}{2}\right)}{\Gamma (a)}\right)^{\frac{1}{a}}=\sqrt{\frac{e}{2a}} \left( 1+\frac{\log (2)}{2 a}+O\left(\frac{1}{a^2}\right)\right)$$

$$\left(\frac{\Gamma \left(\frac{a}{2}\right)}{\Gamma (a)}\right)^{\frac{1}{a}}>\sqrt{\frac{e}{2a}} \qquad \forall a >0$$

Answer 2

Here is a simple approximation, not a strict bound, for the more general form $\left(\dfrac{\Gamma(ca)}{\Gamma(a)}\right)^{1/a}$ which agrees with Claude Leibovici's bound.

For large $a$ and fixed $c > 0$, (this problem is for $c=\frac12$)

$\begin{array}\\ (\Gamma(a))^{1/a} \approx \dfrac{a}{e}\\ \text{so}\\ (\Gamma(ca))^{1/(ca)} &\approx \dfrac{ca}{e}\\ \text{so}\\ (\Gamma(ca))^{1/a} &=((\Gamma(ca))^{1/(ca)})^{c}\\ &\approx (\dfrac{ca}{e})^{c}\\ \text{so}\\ \left(\dfrac{\Gamma(ca)}{\Gamma(a)}\right)^{1/a} &=\dfrac{(\Gamma(ca))^{1/a}}{(\Gamma(a))^{1/a}}\\ &\approx\dfrac{(\dfrac{ca}{e})^{c}}{\dfrac{a}{e}}\\ &=c^c(\dfrac{a}{e})^{c-1}\\ &=c^c(\dfrac{e}{a})^{1-c}\\ \end{array} $

Answer 3

Let $a = 2b$ with $b \ge 1$ an integer. Then \begin{align*} \frac{{\Gamma (b)}}{{\Gamma (2b)}} & = \frac{1}{{(2b - 1)(2b - 2) \cdots (b + 1)b}} \ge \frac{1}{{(2b)(2b) \cdots (2b)(2b)}} \\ & = \frac{1}{{(2b)^b }} \ge \frac{1}{{(2b)^{2b} }} \ge \frac{1}{{(2 \cdot 2b)^{2b} }}. \end{align*} Now let $a = 2b+1$ with $b \ge 2$ an integer. Then \begin{align*} \frac{{\Gamma (b + 1/2)}}{{\Gamma (2b + 1)}} &\ge \frac{{\Gamma (b)}}{{\Gamma (2b + 1)}} = \frac{1}{{2b(2b - 1)(2b - 2) \cdots (b + 1)b}} \\ & \ge \frac{1}{{(2b + 1)(2b + 1) \cdots (2b + 1)(2b + 1)}} = \frac{1}{{(2b + 1)^{b + 1} }} \\ &\ge \frac{1}{{(2b + 1)^{2b + 1} }} \ge \frac{1}{{(2 \cdot (2b + 1))^{2b + 1} }}, \end{align*} where in the first step we used the monotonicity of the gamma function for $b\ge 2$. The case $b=3$ is easy to check by hand.