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?
$S_n$ 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 $\alpha<3$.
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 $b > 3$.
Your expression $S_n$ equals $$\sum_{l=1}^n l\cdot \#\{i \in \mathbb{N},[\sqrt i]= l\}.$$
You should continue as an exercise :-)
