<s>[INST] How closely can we estimate $\sum_{i=0}^n \sqrt{i}$ By looking at an integral and bounding the error?
 [/INST] This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.
The estimate I got was
$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$
for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).
(By $a_{n} \sim b_{n}$ I mean $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0$)
I believe here is a completely elementary proof of that fact:
First consider the inequality for $x > 0$ and $k > 0$.
$$ \sqrt{x} \le \frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}$$
This follows easily by the arithmetic mean $\ge$ geometric mean inequality.
Thus we have that
$$\int_{k}^{k+1} \sqrt{x} \ dx \le \int_{k}^{k+1} (\frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}) \ dx$$
$$ = \frac{(k+1)^2 - k^2}{4\sqrt{k}} + \frac{\sqrt{k}}{2} = \sqrt{k} + \frac{1}{4\sqrt{k}}$$
Thus $$\sum_{k=1}^{n-1} \int_{k}^{k+1} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
i.e.
$$\int_{1}^{n} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
and so
$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
Now we have inequality
$$\frac{1}{2\sqrt{k}} < \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sqrt{k} -\sqrt{k-1}$$
And so, we have that
$$ \sum_{k=1}^{n-1} \frac{1}{4\sqrt{k}} < \frac{\sqrt{n-1}}{2}$$
Thus,
$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \frac{\sqrt{n-1}}{2} + \sum_{k=1}^{n-1} \sqrt{k} $$
So if $$S_{n} = \sum_{k=1}^{n} \sqrt{k}$$ we have that
$$S_{n} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \sqrt{n} - \frac{\sqrt{n-1}}{2} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \frac{\sqrt{n}}{2}$$
Now let $$G_{n} = S_{n} - \frac{2}{3} n^{3/2} - \frac{\sqrt{n}}{2}$$
We have that $$G_{n} \ge -\frac{2}{3}$$
We can easily show that (using tedious but not too complicated algebra*) $$G_{n+1} < G_{n}$$ and so $G_{n}$ is a convergent sequence, as it is bounded below and monotonically decreasing.
Thus there exists a constant $C$ (the limit of $G_{n}$) such that
$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$

* For the sake of completeness, we show that $G_{n+1} < G_{n}$.
Consider $$6(G_{n}-G_{n+1}) = 4(n+1)\sqrt{n+1} + 3\sqrt{n+1} - 6\sqrt{n+1} - 4n\sqrt{n} - 3\sqrt{n}$$
$$ = \sqrt{n+1} + 4n(\sqrt{n+1} -\sqrt{n}) - 3\sqrt{n}$$
Multiplying by $\sqrt{n+1} + \sqrt{n}$ does not change the sign, so we look at
$$ \sqrt{n+1}(\sqrt{n+1} + \sqrt{n}) + 4n - 3\sqrt{n}(\sqrt{n+1} + \sqrt{n})$$
$$ = 2n+1 - 2\sqrt{n^2 + n}$$
Now $$ (2n+1)^2 = 4n^2 + 4n + 1 > 4n^2 + 4n = (2\sqrt{n^2+n}) ^2$$
Hence
$$ 2n+1 > 2\sqrt{n^2+n}$$
and so
$$G_{n} > G_{n+1}$$
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