I'm not sure which would be the best way to compute this limit. As you might have observed, if you expand the infinite sum and rearrange some terms you get:
$$\lim_{n\to \infty}\frac{\sqrt n}{n}\!\cdot\!\sum_{k=0}^n \sqrt k$$
it is easy to see that when you take the limit it yields to an indetermination of the type: $$\mathrm{0}\!\cdot\!\mathrm{\infty}$$ which you can easily transform into one of the type: $$\frac{\infty}{\infty}$$ by dividing by the opposite:
$$\lim_{n\to \infty}\frac{\sum_{k=0}^n \sqrt k}{\frac{n}{\sqrt n}}=\lim_{n\to \infty}\frac{\sum_{k=0}^n \sqrt k}{\sqrt n} $$
I don't know if using L'Hôpital is valid here (as there is an infinite sum in the numerator), how ever, if you differentiate both numerator and denominator, all the square roots in the numerator will go away (except for $\sqrt n$) because they are just constants and you will be left with:
$$\lim_{n\to \infty}\frac{\frac{1}{\sqrt n}}{\frac{1}{\sqrt n}} = \lim_{n\to \infty}\frac{\sqrt n}{\sqrt n} =\lim_{n\to \infty} 1 = 1$$
Is this result correct? I don't even know if it is possible to take the limit of an infinite sum which is not an integral as you are taking discreet steps instead continuous ones.
Thank you for everything.
Best Answer
$\displaystyle{\sum_{k=1}^{n}}\frac{\sqrt k}{\sqrt n}\geqslant n\dfrac1{\sqrt n}=\sqrt n\quad$ for any $\,n\in\Bbb N\,.$
Now take the limit as $\,n\to\infty\,.$