An estimate on the volume of the unit ball Let $d$ be a positive integer. We write $v_d$ for the volume of the $d$-dimensional unit ball.
We know that
\begin{align*}
v_d=\frac{\pi^{d/2}}{\Gamma(d/2+1)}.
\end{align*}
Here, $\Gamma$ denotes the gamma function. That is,
\begin{align}
\Gamma(d/2+1)=\int_{0}^{\infty}t^{d/2}e^{-t}\,dt.
\end{align}
I would like to know an upper estimate for
\begin{align*}
\sup_{d \in \mathbb{N}}\,d\times v_d.
\end{align*}
It easily follows that
\begin{align*}
1/v_d&=\int_{0}^{\infty}(t/\pi)^{d/2}e^{-t}\,dt=\int_{0}^{\infty}u^{d/2} \pi e^{-\pi u}\,du=(d/2)\times \int_{0}^{\infty}u^{d/2-1} e^{-\pi u}\,du.
\end{align*}
If $d\ge 2$, we have
\begin{align*}
\int_{0}^{\infty}u^{d/2-1} e^{-\pi u}\,du \ge \int_{1}^{\infty} e^{-\pi u}\,du =(1/\pi)\times e^{-\pi}.
\end{align*}
Therefore, we obtain that
\begin{align*}
d \times v_d \le 2\pi \times e^\pi.
\end{align*}
However, I don't think this bound is  sharp. Can you give a better estimate (I want to find it in the simplest way possible)?
 A: Well, you could simply calculate the first few values and find the maximum (incidentally, it occurs at $d=7$). Then prove, for example by considering the ratios $u_{d+2}/u_d$ where $u_d = dv_d$, that later values cannot exceed that, so then you have a sharp upper bound because you know the maximum.
I don't see how it could be made much simpler, because it really boils down to how the $\Gamma$ function in the denominator eventually grows quicker than the $d \pi^{d/2}$ in the numerator, so you basically need to observe when that happens. Of course you could replace the $\Gamma$ with some crude lower bound, but I don't see much advantage here over the first option.
A: $$d \times V_d=\frac{d \,\pi ^{d/2}}{\Gamma \left(\frac{d}{2}+1\right)}$$
$$\big[d \times V_d\big]'=\frac{\pi ^{d/2}}{d \,\Gamma \left(\frac{d}{2}\right)}\Bigg[d\, \log (\pi )-d \,\psi\left(\frac{d}{2}+1\right)+2 \Bigg]$$ Using asymptotics
$$d\, \log (\pi )-d \,\psi\left(\frac{d}{2}+1\right)+2 =d \log \left(\frac{2 \pi }{d}\right)+1+\frac{1}{3 d}+O\left(\frac{1}{d^3}\right)$$
$$d \log \left(\frac{2 \pi }{d}\right)+1=0 \implies d=\frac{1}{W\left(\frac{1}{2 \pi }\right)}=7.21700$$ while the exact solution would be $7.25695$
Checking for $d=7$ we have $d \times V_d=\frac{16 \pi ^3}{15}=33.07$ while, for $d=8$, we have $d \times V_d=\frac{\pi ^4}{3}=32.47$.
