<s>[INST] how to compute $\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n$? I need to compute $$\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n dx.$$
I proved that
for $n\ge1$,
$$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!},$$
but I don't know how to continue from here.
I also need to calculate $\int_{0}^{1}(1-x^2)^ndx$ for $n=50$ with a $1$% accuracy. I thought about using Taylor series but also failed.
 [/INST] Redoing what has been done
many many many times before.
$\begin{array}\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
I_0
&=\int_0^1  dx\\
&= 1\\
I_1
&=\int_0^1 (1-x^2) dx\\
&=1-\dfrac13\\
&=\dfrac23\\
I_n
&=\int_0^1 (1-x^2)^n dx\\
&=x(1-x^2)^n|_0^1+\int_0^1 2x^2n(1-x^2)^{n-1} dx\\
&\qquad\text{integrating by parts}\\
&\qquad f = (1-x^2)^n, f' = -2xn(1-x^2)^{n-1}, g' = 1, g = x\\
&=2n\int_0^1 x^2(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (x^2-1+1)(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-(1-x^2))(1-x^2)^{n-1} dx\\
&=2n\int_0^1 (1-x^2)^{n-1} dx-2n\int_0^1 (1-x^2)^{n} dx\\
&=2nI_{n-1}-2nI_n\\
\text{so}\\
I_n
&=\dfrac{2n}{2n+1}I_{n-1}\\
\dfrac{I_n}{I_{n-1}}
&=\dfrac{2n}{2n+1}\\
I_n
&=\dfrac{I_n}{I_{0}}\\
&=\prod_{k=1}^n\dfrac{I_k}{I_{k-1}}\\
&=\prod_{k=1}^n\dfrac{2k}{2k+1}\\
&=\dfrac{\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k+1)}\\
&=\dfrac{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k)}{\prod_{k=1}^n(2k)\prod_{k=1}^n(2k+1)}\\
&=\dfrac{4^nn!^2}{(2n+1)!}\\
&=\dfrac{4^nn!^2}{(2n)!(2n+1)}\\
&\approx\dfrac{4^n(\sqrt{2\pi n}(n/e)^n)^2}{\sqrt{2\pi 2n}(2n/e)^{2n}(2n+1)}
\qquad\text{Stirling strikes twice}\\
&=\dfrac{4^n((2\pi n)(n^{2n}/e^{2n})}{2\sqrt{\pi n}4^nn^{2n}e^{2n}(2n+1)}\\
&=\dfrac{2\pi n}{2\sqrt{\pi n}(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{(2n+1)}\\
&=\dfrac{\sqrt{\pi n}}{2n(1+1/(2n))}\\
&=\dfrac{\sqrt{\pi n}}{2n}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}\dfrac1{1+1/(2n)}\\
&=\dfrac{\sqrt{\pi }}{2\sqrt{n}}(1-\dfrac1{2n}+O(\dfrac1{n^2}))\\
\end{array}
$
so
$\sqrt{n}I_n
=\dfrac{\sqrt{\pi }}{2}(1-\dfrac1{2n}+O(\dfrac1{n^2}))
\to\dfrac{\sqrt{\pi }}{2}
$.
</s>