Show that for $p \geq 1$, $[\Gamma(\frac{p+1}{2})]^{1/p}=O(\sqrt{p})$ as $p \rightarrow \infty$ 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!
 A: 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
$$ \Gamma\left(\frac{p+1}{2}\right)^{\frac{1}{p}}\underset{p\rightarrow +\infty}{\sim}\sqrt{\frac{p}{2e}} $$
A: $$
\begin{align}
\Gamma\!\left(\frac{p+1}2\right)
&=\Gamma\!\left(1+\left\{\frac{p-1}2\right\}\right)\prod_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\tag1\\
&\le1\cdot\left[\frac1{\left\lfloor\scriptstyle{\frac{p-1}2}\right\rfloor}\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\right]^{\left\lfloor\frac{p-1}2\right\rfloor}\tag2\\
&=\left(\frac{p+1}4+\frac12\left\{\frac{p+1}2\right\}\right)^{\left\lfloor\frac{p-1}2\right\rfloor}\tag3
\end{align}
$$
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 $p\ge1$, $\left(\frac{p+3}{4p}\right)^p\le1\le\frac{p+3}4\implies\left(\frac{p+3}4\right)^{\frac{p-1}{2p}}\le\sqrt{p}$

A: $\Gamma$ 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.
