Inequality by telescope It can be shown through less elementary methods that:
$\displaystyle \large \frac{2}{3}n\sqrt{n} < \sum_{k=1}^n \sqrt{k} < \frac{2}{3}n\sqrt{n} + \frac{\sqrt{n}}{2}$
It is easy, but tedious, through squaring, to show that:
Lemma:
$\displaystyle \large \frac{3}{2}\sqrt{k} > \left(k+\frac{1}{2}\right)^{\frac{3}{2}} - \left(k-\frac{1}{2}\right)^{\frac{3}{2}} $
Then telescope to obtain a slightly stronger lower bound than what we seek and acquire the lower bound with some basic inequalities
$\large 1. $ Does there exist an easier way to prove the lemma that doesn't use calculus or induction? Exploiting monotonicity and convexity does not count as calculus, for the purposes of this question.
$\large 2. $ Is there a telescope for the upper bound? I have tried with $\displaystyle \left(k+\frac{2}{3}\right)^{\frac{3}{2}} - \left(k-\frac{1}{3}\right)^{\frac{3}{2}}, \left(k+\frac{3}{4}\right)^{\frac{3}{2}} - \left(k-\frac{1}{4}\right)^{\frac{3}{2}} \text{ and } \left(k+\frac{3}{5}\right)^{\frac{3}{2}} - \left(k-\frac{2}{5}\right)^{\frac{3}{2}}$
to no success, as the upper bound is stronger than all three obtained from telescoping.

Let $\frac{3}{2}\sqrt{k}=a$, $\sqrt{\left(k+\frac{1}{2}\right)^3}=b$ and $\sqrt{\left(k-\frac{1}{2}\right)^3}=c$.
Since $b>c$, we see that $a+b-c>0$ and $$b+c=\left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}+\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}}\geq\sqrt{k}+1\cdot\frac{1}{2}\sqrt{k}=\frac{3}{2}\sqrt{k}=a,$$
which says, that $b+c-a>0$.
But
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=$$
$$=2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=\frac{1}{16}(3k^2-1)>0,$$
which says that $a+c>b$ and we are done!
