Consider this limit

Question

$\lim_{n\rightarrow\infty}n^k\int_0^n(\sin x\sin\frac x{2^2}...\sin\frac x{n^2})dx=?,\;for\;any\;k\;\in\mathbb{N}^\ast$

I tried to work the inside product of the integral out.. but couldn't get anywhere...

Answer 1

For any $x\in[0,n]$ we have $$0\leq \prod_{h=1}^{n}\sin\frac{x}{h^2} \leq \prod_{h=1}^{n}\frac{x}{h^2} = \frac{x^n}{n!^2} $$ hence $$ 0\leq \int_{0}^{n}\prod_{h=1}^{n}\sin\frac{x}{h^2}\,dx \leq \int_{0}^{n}\frac{x^n}{n!^2}\,dx = \frac{n^{n+1}}{(n+1)n!^2}$$ and for any $k\in\mathbb{N}$ we have $$ \lim_{n\to +\infty}\frac{n^{n+k+1}}{(n+1)n!^2} = \color{red}{0}.$$