Show that for $p \geq 1$, $[\Gamma(\frac{p+1}{2})]^{1/p}=O(\sqrt{p})$ as $p \rightarrow \infty$

Question

Actually it is Exercise 2.5.1 in High Dimensional Probability by Vershynin.

I got stuck at showing that for $p \geq 1$, $$\left[\Gamma\left(\frac{p+1}{2}\right)\right]^{1/p}=O(\sqrt{p}) \ \ \ \text{as }p\rightarrow\infty$$ I can show it if $p\in \mathbb{N}$ but I have no idea for real $p\geq 1$.

Any kinds of help is appreciated, thank you!

Answer 1

Using Stirling's formula, we have $$ \Gamma\left(\frac{p+1}{2}\right)=\sqrt{\pi p}\left(\frac{p+1}{2e}\right)^{\frac{p+1}{2}}(1+o(1)) $$ Thus $$ \Gamma\left(\frac{p+1}{2}\right)^{\frac{1}{p}}=\left(\frac{p+1}{2e}\right)^{\frac{p+1}{2p}}(\pi p+o(p))^{\frac{1}{2p}} $$ because $\pi p(1+o(1))^2=\pi p+o(p)$. $$ \begin{aligned}\left(\frac{p+1}{2e}\right)^{\frac{p+1}{2p}}&=\exp\left(\frac{p+1}{2p}\ln\left(\frac{p+1}{2e}\right)\right) \\ &=\exp\left(\frac{p+1}{2p}\ln(p+1)-\frac{p+1}{2p}\ln(2e)\right) \\ &=\exp\left(\frac{\ln p}{2}-\frac{\ln 2+1}{2}+o(1)\right) \\ &=\sqrt{\frac{p}{2}}e^{-\frac{1}{2}+o(1)} \end{aligned} $$ and $$ (\pi p+o(p))^{\frac{1}{2p}}=\exp\left(\frac{1}{2p}\ln(\pi p+o(p))\right)=e^{o(1)} $$ Thus $$ \Gamma\left(\frac{p+1}{2}\right)^{\frac{1}{p}}\underset{p\rightarrow +\infty}{\sim}\sqrt{\frac{p}{2e}} $$

Answer 2

$$ \begin{align} \Gamma\!\left(\frac{p+1}2\right) &=\Gamma\!\left(1+\left\{\frac{p-1}2\right\}\right)\prod_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\tag1\\ &\le1\cdot\left[\frac1{\left\lfloor\scriptstyle{\frac{p-1}2}\right\rfloor}\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\right]^{\left\lfloor\frac{p-1}2\right\rfloor}\tag2\\ &=\left(\frac{p+1}4+\frac12\left\{\frac{p+1}2\right\}\right)^{\left\lfloor\frac{p-1}2\right\rfloor}\tag3 \end{align} $$ Explanation:
$(1)$: use the recurrence formula $\Gamma(x+1)=x\,\Gamma(x)$
$(2)$: $\Gamma(x)\le1$ for $1\le x\le2$ and the AM-GM inequality
$(3)$: evaluate the sum

Therefore, $$ \begin{align} \Gamma\left(\frac{p+1}2\right)^{1/p} &\le\left(\frac{p+3}4\right)^{\frac{p-1}{2p}}\tag4\\[6pt] &\le\sqrt{p}\tag5 \end{align} $$ Explanation:
$(4)$: inequality $(3)$ and $0\le\left\{\frac{p+1}2\right\}\lt1$
$(5)$: for $p\ge1$, $\left(\frac{p+3}{4p}\right)^p\le1\le\frac{p+3}4\implies\left(\frac{p+3}4\right)^{\frac{p-1}{2p}}\le\sqrt{p}$

Answer 3

$\Gamma$ is log-convex by the Bohr-Mollerup theorem or just from the integral representation and Cauchy-Schwarz, so $$\Gamma\left(\frac{p+1}{2}\right)\leq \sqrt{\Gamma(1)\Gamma(p)}=\sqrt{\Gamma(p)}\leq \sqrt{\Gamma(p+1)}\leq\sqrt{p^p} $$ and the claim is trivial.