TITLE: On a sum involving fractional part of an integer
QUESTION [5 upvotes]: I was interested in estimating the sum of the form
$$
\sum_{j=1}^{N} \{ \sqrt{j} \}.
$$
I was wondering if there is a reference or maybe some one could
help me figure out what to do.
Thanks!
$\{ \alpha\}$ denotes the fractional part of the real number $\alpha$.

REPLY [3 votes]: The desired sum is equal to $$\left(\sum_{j=1}^N \sqrt{j}\right) - \left(\sum_{j=1}^N \lfloor \sqrt{j}\rfloor\right)$$
The second sum is computed here as $(N+1)a - \frac{a^3}{3} - \frac{a^2}{2} - \frac{a}{6}$, where $a=\lfloor \sqrt{N+1}\rfloor$.
The first sum may be estimated by an integral as follows
$$\frac{2}{3}N^{3/2}=\int_0^N\sqrt{x}dx < \left(\sum_{j=1}^N \sqrt{j}\right) < \int_1^{N+1} \sqrt{x}dx=\frac{2}{3}((N+1)^{3/2}-1)$$
The difference between the upper and lower estimates is $O(\sqrt{N})$, which isn't too bad I think.  With care the integrals can be refined, sharpening the estimates.