\begin{align} \lim_{n \to \infty} \frac{n^{\frac{3}{2}} a_n}{4^n} &= \frac{e}{\sqrt{\pi}} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{-n-\frac{3}{2}} \cdot \frac{(2n)! e^{2n}}{(2n)^{2n} \sqrt{4 \pi n}} \cdot \frac{(n+1)^{n+1} \sqrt{2 \pi (n+1)}}{(n+1)! e^{n+1}} \cdot \frac{n^n \sqrt{2 \pi n}}{n! e^n} \\ &= \frac{e}{\sqrt{\pi}} \cdot e^{-1} \cdot 1 \cdot 1 \cdot 1 \\ &= \frac{1}{\sqrt{\pi}} \end{align}
$t=\sqrt{1-x}$ とおくと、 $x=1-t^2, \ dx=-2tdt$ であり、次のように計算できる： \begin{align} \int_{\frac{1}{2}}^1 \frac{dx}{x \sqrt{1-x}} &= \int_{\frac{1}{\sqrt{2}}}^0 \frac{-2tdt}{(1-t^2)t} \\ &= \int_0^{\frac{1}{\sqrt{2}}} \frac{-2dt}{(t+1)(t-1)} \\ &= \int_0^{\frac{1}{\sqrt{2}}} \left( \frac{1}{t+1} - \frac{1}{t-1} \right) dt \\ &= \left[ \log \left| t+1 \right| - \log \left| t-1 \right| \right]_0^{\frac{1}{\sqrt{2}}} \\ &= \log \frac{\sqrt{2}+1}{\sqrt{2}} - \log \frac{\sqrt{2}-1}{\sqrt{2}} \\ &= \log \frac{\sqrt{2}+1}{\sqrt{2}-1} \\ &= \log \left( 3 + 2 \sqrt{2} \right) \end{align}