We have the sequence {$I_n$} where $n\in\Bbb N$ and $$I_n=\int_0^{\pi\over 2} e^{-x}\sin^nx\;dx$$ It can be proven that $$(n^2+1)I_n=-e^{-{\pi\over 2}}+n(n-1)I_{n-2}$$for $n\geq 2$.
How to show that as $n\to \infty, I_n\to 0$? The problem is that if we assume $I_n\to I$, we only get $I\to 0$ instead of $I=0$. Is this method always correct?
Note that since $e^{-x} \le 1$ on $\left[0,\frac \pi2\right]$, for all $n$, $$0 < I_n < J_n = \int_0^{\frac\pi 2}\sin^n x\,dx$$
$J_n$ has a nicer recursion formula from which it is easy to prove $J_n \to 0$, and the squeeze theorem takes care of the rest.