Given $s_{n}=\left\{\begin{matrix} 0 & &x=0 \\ \sqrt{n} & & x\in\left [ \frac{1}{n+1},\frac{1}{n} \right ] \end{matrix}\right.$, show that $\left\{\int_{0}^{1}s_{n}\right\}$ is bounded above.
I would think it is because as $n\to\infty$, isn't $s_{n}\to 0$?
$\int_0^{1} s_n =\sqrt n (\frac 1 n -\frac 1{n+1})=\sqrt n \frac 1 {n(n+1)} \in [0,1]$.