$\sqrt{k} - 1 < \left \lfloor \sqrt{k} \right \rfloor \leq \sqrt{k}$ Hence, $\sum_{k=1}^{n} \left(\sqrt{k} - 1 \right) < \sum_{k=1}^{n} \left(\left \lfloor \sqrt{k} \right \rfloor \right) \leq \sum_{k=1}^{n} \left(\sqrt{k} \right)$
Now us the fact that $\sum_{k=1}^n \sqrt{k} = \dfrac23 n^{3/2} + \dfrac{\sqrt{n}}2 + \dfrac1{24} \dfrac1{\sqrt{n}} + \zeta(-1/2) + \mathcal{O}(1/n)$
$\dfrac23 n^{3/2} - n + \mathcal{O}(n^{1/2}) < \sum_{k=1}^{n} \left(\left \lfloor \sqrt{k} \right \rfloor \right) \leq \dfrac23 n^{3/2} + \mathcal{O}(n^{1/2})$
$\dfrac23 n^{3} - n^2 + \mathcal{O}(n) < \sum_{k=1}^{n^2} \left(\left \lfloor \sqrt{k} \right \rfloor \right) \leq \dfrac23 n^{3} + \mathcal{O}(n)$