$S_n=\sum_{k=1}^{n^2} [\sqrt{k}]$ 
Let $$S_n=\sum_{k=1}^{n^2} [\sqrt{k}]$$where $[x]$ represents the floor function. Find the limit of the sequence $(a_n)_{n\geq1}$, where $$a_n=\frac{S_n}{n^\alpha}$$where $\alpha\in\mathbb{R}$

After applying Cesaro-Stolz we obtain that $$\lim_{n\to+\infty}\frac{ \sum_{k=n^2+1}^{(n+1)^2} [\sqrt{k}]}{(n+1)^\alpha-n^\alpha}$$ and I wonder if there is any formula that helps us getting away from that sum, or maybe we shouldn't even apply Cesaro-Stolz. Any idea?
 A: $S_n$ is the number of lattice points in the region $\{(x,y):0<x\leq n^2, 0<y\leq\sqrt{x}\}$, which is half a parabolic segment with area $\frac{2}{3}n^3$ and perimeter $O(n^2)$. In particular (like in the Gauss circle problem) elementary arguments give $S_n=\frac{2}{3}n^3+O(n^2)$ and $\lim_{n\to +\infty}\frac{S_n}{n^\alpha}$ equals $\frac{2}{3}$ for $\alpha=3$, zero for $\alpha>3$ and $+\infty$ for $\alpha<3$.
A: If
$T_n
=\sum_{k=1}^{n^2} \sqrt{k}
$
then
$0
\lt T_n-S_n
=\sum_{k=1}^{n^2}( \sqrt{k}- [\sqrt{k}])
\lt n^2
$
or
$T_n-n^2
\lt S_n
\lt T_n
$.
(What follows is very standard.)
Since,
for $a > 0$
we have
$\int_{k-1}^k x^{a} dx
\lt k^a
\lt \int_{k}^{k+1} x^{a} dx
$,
$\sum_{k=1}^m k^a
\gt \sum_{k=1}^m \int_{k-1}^k x^{a} dx
= \int_{0}^m x^{a} dx
=\dfrac{m^{a+1}}{a+1}
$
and
$\sum_{k=1}^m k^a
=m^a+\sum_{k=1}^{m-1} k^a
\lt m^a+\sum_{k=1}^{m-1} \int_{k}^{k+1} x^{a} dx
= m^a+\int_{1}^m x^{a} dx
\lt m^a+\dfrac{m^{a+1}}{a+1}
$,
we have
$0
\lt \sum_{k=1}^m k^a-\dfrac{m^{a+1}}{a+1}
\lt m^a
$.
Setting $a=\frac12$
and $m = n^2$,
$0
\lt T_n-\dfrac{n^{3}}{3/2}
\lt n
$
or
$\dfrac{2n^{3}}{3}
\lt T_n
\lt \dfrac{2n^{3}}{3}+n
$.
Putting this
in the original inequality
for $S_n$,
we have
$\dfrac{2n^{3}}{3}-n^2
\lt S_n
\lt \dfrac{2n^{3}}{3}+n
$
or
$\dfrac{2}{3}-\dfrac1{n}
\lt \dfrac{S_n}{n^3}
\lt \dfrac{2}{3}+\dfrac1{n^2}
$.
Therefore
$\lim_{n \to \infty} \dfrac{S_n}{n^b}
=\infty$ if $b < 3$,
$=\dfrac23$
if $b=3$,
and
$=0$
of $b > 3$.
A: Your expression $S_n$ equals
$$\sum_{l=1}^n l\cdot \#\{i \in \mathbb{N},[\sqrt i]= l\}.$$
You should continue as an exercise :-)
