Proof: $2\sqrt{m}-2 < \sum\limits_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1$

Well you tagged induction so why not use it:

$$ \sum_{n=1}^{m+1} \frac{1}{\sqrt{n}} \lt 2\sqrt{m}-1 + \frac{1}{\sqrt{m+1}} \lt 2\sqrt{m}-1 + 2(\sqrt{m+1}-\sqrt{m}) = 2\sqrt{m+1}-1 $$