Limit of a Lebesgue integral What is the value of:
$$\lim_{n\to\infty}\sqrt{n}\int_0^{1}(1-t^2)^ndt$$
I think I have to use the Theorem of dominated convergence 
 A: You can find the value of the limit by squeezing, since:
$$\sqrt{n}\int_{0}^{1}(1-t^2)^n dt\leq \sqrt{n}\int_{0}^{1}e^{-nt^2}dt = \int_{0}^{\sqrt{n}}e^{-t^2}dt<\int_{0}^{+\infty}e^{-t^2}dt=\frac{\sqrt{\pi}}{2},$$
and the differences between the first and the second term, the third and the fourth, are $O\left(\frac{1}{\sqrt{n}}\right).$
Through the $t=\cos\theta$ substitution you can also recognize the Wallis product in the LHS.
A: From my comment:
$$\begin{align*}
A_n & = \int_0^1 (1-t^2)^n dt = \int_0^1 (1-t^2)^{n-1} dt - \int_0^1 t^2 (1-t^2)^{n-1} dt \\
& = A_{n-1} - \left( \underbrace{\frac{1}{n} \left[ t^2 (1-t^2)^n \right]_0^1}_{=0} - \frac{2}{n} \int_0^1 t (1-t^2)^n dt \right) \\
& = A_{n-1} + \frac{2}{n} \left( \underbrace{\frac 1 {n+1} \left[ t(1-t^2)^{n+1} \right]_0^1}_{=0} - \frac{1}{n+1} \int_0^1 (1-t^2)^{n+1} \right) \\
& = A_{n-1} - \frac 2 {n(n+1)} A_{n+1}
\end{align*}$$
So
$$A_{n+1} = \frac{n(n+1)} 2 (A_{n-1} - A_n)$$
With $A_0 = 1, A_1 = \frac{2}{3}$.
Does this get you any further?
