$Prove for $n\in \mathbb{N}$, that $\sqrt{n} \leq \sum_{r=1}^{n}\frac{1}{\sqrt{r}} \leq 2\sqrt{n}-1$.
Not sure how to do this. Tried splitting the inequality up into \sqrt{n} \leq \sum_{r=1}^{n}\frac{1}{\sqrt{r}} and \sum_{r=1}^{n}\frac{1}{\sqrt{r}} \leq 2\sqrt{n}-1 and then prove these...