<s>[INST] How closely can we estimate $\sum_{i=0}^n \sqrt{i}$ By looking at an integral and bounding the error?
 [/INST] In case you were wondering, like me, Moron's excellent proof adapts easily to show that
$$1 + \sqrt[3]{2} + \dots + \sqrt[3]{n} \sim \frac{3}{4}n^{4/3} + \frac{\sqrt[3]{n}}{2} + C,$$
for some constant $C.$ In this case $C = \zeta(-1/3) \approx -0.277343.$
Where, as before, $a_n \sim b_n$ means $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0.$
Similar to the previous proof, we use the AM-GM inequality to show
$$\sqrt[3]{x} \le \frac{x}{3k^{2/3}} + \frac{2k^{1/3}}{3}.$$
Summing from $k=1$ to $n-1$ and integrating we arrive at
$$ \frac{3}{4}n^{4/3} - \frac{3}{4} \le \sum_{k=1}^{n-1}\sqrt[3]{n} +
\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$
And so,
$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + n^{1/3} - \frac{3}{4} - \frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}}, \qquad n>1.$$
Using $\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le 1+ \int_1^n x^{-2/3} dx,
\qquad n>1,$ we obtain
$$\frac{1}{6}\sum_{k=1}^{n-1} \frac{1}{k^{2/3}} \le \frac{1}{2}n^{1/3} - \frac{1}{3}.$$
And hence
$$\sum_{k=1}^{n}\sqrt[3]{n} \ge \frac{3}{4}n^{4/3} + \frac{1}{2}n^{1/3} - \frac{5}{12}, \qquad n>1.$$
As in the previous argument, set
$$G_n = \sum_{k=1}^{n}\sqrt[3]{n} - \frac{3}{4}n^{4/3} - \frac{1}{2}n^{1/3}.$$
Then $G_n \ge -5/12,$ and we can show $G_n – G_{n+1} > 0$ by showing that
$$\frac{(n+1)^{4/3} – n^{4/3}}{(n+1)^{1/3} + n^{1/3}} > \frac{2}{3}.$$
So, as before, $G_n$ is convergent, since it is bounded below and monotonically decreasing.
I suspect this argument also adapts easily to the more general case
$\sum_{k=1}^{n}\sqrt[r]{n},$ for $r \in \mathbb{N},$ where I'm guessing, we'll find
$$1 + \sqrt[r]{2} + \dots + \sqrt[r]{n} \sim \frac{r}{r+1}n^{(r+1)/r} + \frac{\sqrt[r]{n}}{2} + C,$$ 
where $C = \zeta(-1/r).$ However, I confess to not having checked the details of this further generalisation.
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