How can I prove that the following sequence is convergent? The sequence is:
$$
I(n) = \int_{0}^{\pi/2}\cos^n(x) dx, \; n\geq 1
$$
I just proved that it is a monotone decreasing sequence, but I don't know what to do next. I'm pretty sure that's not all I need to do to prove it is convergent.
 A: Denoting $ \left(\forall n\in\mathbb{N}\right),\ W_{n}=\displaystyle\int_{0}^{\frac{\pi}{2}}{\cos^{n}{x}\,\mathrm{d}x} : $
We have : \begin{aligned} \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=\displaystyle\int_{0}^{\frac{\pi}{2}}{\cos{x}\cos^{n}{x}\,\mathrm{d}x} \\ &=\left[\sin{x}\cos^{n}{x}\right]_{0}^{\frac{\pi}{2}}+n\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin^{2}{x}\cos^{n-1}{x}\,\mathrm{d}x}\\ &=n\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(1-\cos^{2}{x}\right)\cos^{n-1}{x}\,\mathrm{d}x}\\ \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=n\left(W_{n-1}-W_{n+1}\right)\\ \iff \left(\forall n\in\mathbb{N}^{*}\right),\ W_{n+1}&=\displaystyle\frac{n}{n+1}W_{n-1} \end{aligned}
And since $ \left(W_{n}\right)_{n\in\mathbb{N}} $ is positive and decreasing, we have that : $$ \left(\forall n\geq 2\right),\ W_{n+1}\leq W_{n}\leq W_{n-1}\iff \displaystyle\frac{n}{n+1}\leq\displaystyle\frac{W_{n}}{W_{n-1}}\leq 1 $$
Thus $ \displaystyle\lim_{n\to +\infty}{\displaystyle\frac{W_{n}}{W_{n-1}}}=1 \cdot $
We can easily verify that the sequence $ \left(y_{n}\right)_{n\in\mathbb{N}} $ defined as following $ \left(\forall n\in\mathbb{N}\right),\ y_{n}=\left(n+1\right)W_{n}W_{n+1} $ is a constant sequence. (Using the recurrence relation that we got from the integration by parts to express $ W_{n+1} $ in terms of $ W_{n-1} $ will solve the problem)
Hence $ \left(\forall n\in\mathbb{N}\right),\ y_{n}=y_{0}=W_{0}W_{1}=\displaystyle\frac{\pi}{2} \cdot $
Now that we've got all the necessary tools, we can prove that $ \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}}=\sqrt{\displaystyle\frac{\pi}{2}} : $ \begin{aligned} \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}} &=\displaystyle\lim_{n\to +\infty}{\sqrt{y_{n-1}}\sqrt{\displaystyle\frac{W_{n}}{W_{n-1}}}}\\ &=\displaystyle\lim_{n\to +\infty}{\sqrt{\displaystyle\frac{\pi}{2}}\sqrt{\displaystyle\frac{W_{n}}{W_{n-1}}}}\\ \displaystyle\lim_{n\to +\infty}{\sqrt{n}W_{n}}&=\sqrt{\displaystyle\frac{\pi}{2}} \end{aligned}
From that we can get that : $$ \lim_{n\to +\infty}{W_{n}}=\lim_{n\to +\infty}{\left(\frac{1}{\sqrt{n}}\times\sqrt{n}W_{n}\right)}=0\times\sqrt{\frac{\pi}{2}}=0 $$
A: Let $f_n(x) = \left( 1 - \dfrac{4x^2}{\pi^2}\right)^n$. The inequality $0 \le \cos x \le f_1(x)$ holds on the interval $[0,\pi/2]$ so that $$0 \le \int_0^{\pi/2} \cos^n x \, dx \le \int_0^{\pi/2} f_n(x) \, dx$$ for all $n \ge 1$.
The sequence $f_n(x)$ converges uniformly to $0$ on $[\epsilon,\pi/2]$ for any $\epsilon > 0$. Thus given any $0 < \epsilon < \pi/2$ you can choose $N$ so that $$n \ge N \implies 0 \le f_n(x) < \frac{\epsilon}{\pi/2 - \epsilon}$$
for all $x \in [\epsilon,\pi/2]$.  Consequently
$$n \ge N \implies  0 \le \int_0^{\pi/2} f_n(x) \, dx =  \int_0^{\epsilon} f_n(x) \, dx  + \int_\epsilon^{\pi/2} f_n(x) \, dx < 2\epsilon.$$
