let $p$ be a continuous function on $[0,1]$ and $(u_n)$ a sequence such that : $\sqrt{u_n} = \pi n + O(\frac{1}{n})$ when $n \to \infty$.
Now I need to prove that the sequence : $$v_n = \frac{\sin(\sqrt{u_n})}{\sqrt{u_n}} + \frac{1}{u_n}\int_0^1\sin(\sqrt{u_n}(1-x))p(x)\sin(\sqrt{u_n}x) \mathrm{d}x$$
is an $O(\frac{1}{n^3})$ when $n \to \infty$
I am must say that I don't understand how $v_n$ can be a $O(\frac{1}{n^3})$. I mean we have : $\sin(\sqrt{u_n}) = (-1)^k \sin(O(1/n))$ thus when $n$ goes to infinty $\mid \sin(\sqrt{u_n}) \mid \sim O(1/n)$. Thus we get that : $\mid \frac{\sin(\sqrt{u_n})}{\sqrt{u_n}} \mid \sim O(\frac{1}{n^2})$. So here already I don't get the cube...
Thank you !