TITLE: How closely can we estimate $\sum_{i=0}^n \sqrt{i}$
QUESTION [21 upvotes]: By looking at an integral and bounding the error?

REPLY [2 votes]: By the Euler-Maclaurin summation formula, we have
\begin{align}\sum_{k=1}^n\sqrt k&=\frac23n^{3/2}+\frac12n^{1/2}+\zeta(-1/2)+\sum_{k=1}^\infty\frac{B_{2k}\Gamma(\frac32)}{(2k)!\Gamma(\frac52-2k)}n^{\frac32-2k}\\&=\frac23n^{3/2}+\frac12n^{1/2}+\zeta(-1/2)+\frac1{24}n^{-1/2}+\frac1{1920}n^{-3/2}+\mathcal O(n^{-7/2})\end{align}
Seeing as the remainder term in the expansion goes to zero for $n\ge1$ and the collected constants add up to the zeta function.
A graph of the terms as far as expanded is shown below:

More generally, we have, for $s\ne-1$ and large enough $n$,
$$\sum_{k=1}^nk^s=\zeta(-k)+\frac1{s+1}n^{s+1}+\frac12n^s+\sum_{k=1}^\infty\frac{B_{2k}\Gamma(s+1)}{(2k)!\Gamma(s+1-k)}n^{s-k}$$
As implimented in this graph.