Convergence of a series with integral inside I'm trying to determine whether the series 
$$\sum_{n=0}^{+\infty}\int_0^{\frac{\pi}{2}}\sin^n(x)dx$$
converges, diverges or is indeterminate. 
Since $\sin(x)\le 1\;\forall x \in \mathbb{R}$ and thus $\sin^n(x)\le 1^n=1 \; \forall n \in \mathbb{N}$ I tried with the comparison test but with no luck. Does anyone know how to solve this exercise?
Thanks in advance.
 A: For first, the hard way. Through the Beta function and the Euler product for the $\Gamma$ function:
$$ \int_{0}^{\pi/2}\sin(x)^n\,dx = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\cdot\Gamma\left(1+\frac{n}{2}\right)}\sim\frac{C}{\sqrt{n}}\tag{1}$$
hence the series is not convergent by the $p$-test. The easy way: by assuming that the series is converging, we are allowed to switch the series and the integral, reaching:
$$ \int_{0}^{\pi/2}\frac{dx}{1-\sin x}=\int_{0}^{1}\frac{dx}{(1-x)\sqrt{1-x^2}}=\int_{0}^{1}\frac{dx}{x\sqrt{x(2-x)}}\tag{2}$$
where the last integral is not converging since the integrand function has a non-integrable singularity in a right neighbourhood of zero. The inequality:
$$ \int_{0}^{\pi/2}\frac{dx}{1-\sin x}=\int_{0}^{\pi/2}\frac{dx}{1-\cos x}=\int_{0}^{\pi/2}\frac{dx}{2\sin^2\frac{x}{2}}\geq \int_{0}^{\pi/2}\frac{2\,dx}{x^2}\tag{3}$$
makes it trivial.
A: By drawing a picture, we note that for $0<x<\pi/2$,
$$ \frac{2}{\pi}x < \sin{x} < x. $$
In particular, therefore, we have
$$ \int_0^{\pi/2} \sin^n{x} \, dx > \int_0^{\pi/2} \left( \frac{2}{\pi} x \right)^n \, dx = \frac{1}{n+1}\frac{2}{\pi} \left( \frac{2}{\pi}\frac{\pi}{2} \right)^{n+1} = \frac{2}{\pi(n+1)}. $$
Hence
$$ \sum_{n=0}^{\infty} \int_0^{\pi/2} \sin^n{x} \, dx > \sum_{n=0}^{\infty} \frac{2}{\pi(n+1)}, $$
which of course diverges, so the original sum diverges.
