On the asymptotics of $\sum_{k=1}^{n^2} \{\sqrt{k}\} $

On the asymptotics of $\sum_{k=1}^{n^2} \{\sqrt{k}\} $

This was inspired by a problem in Quora which asked to show that $s(n) =\sum_{k=1}^{n^2} \{\sqrt{k}\} \le \frac{n^2-1}{2} $.

($\{...\}$ means fractional part.)

Getting bounds on $s(n)$ is relatively straightforward:

$\begin{array}\\ s(n) &=\sum_{k=1}^{n^2} \{\sqrt{k}\}\\ &=\sum_{j=1}^{n-1}\sum_{k=j^2}^{j^2+2j} \{\sqrt{k}\}\\ &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \{\sqrt{j^2+k}\}\\ &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} (\sqrt{j^2+k}-j)\\ &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} (\sqrt{j^2+k}-j)\dfrac{\sqrt{j^2+k}+j}{\sqrt{j^2+k}+j}\\ &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \dfrac{k}{\sqrt{j^2+k}+j}\\ &\le\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \dfrac{k}{2j}\\ &=\sum_{j=1}^{n-1}\dfrac1{2j}\sum_{k=0}^{2j} k\\ &=\sum_{j=1}^{n-1}\dfrac1{2j}\dfrac{2j(2j+1)}{2}\\ &=\dfrac12\sum_{j=1}^{n-1}(2j+1)\\ &=\dfrac{n^2-1}{2}\\ s(n) &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \dfrac{k}{\sqrt{j^2+k}+j}\\ &\gt\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \dfrac{k}{2j+1}\\ &=\sum_{j=1}^{n-1}\dfrac1{2j+1}\sum_{k=0}^{2j} k\\ &=\sum_{j=1}^{n-1}\dfrac1{2j+1}\dfrac{2j(2j+1)}{2}\\ &=\sum_{j=1}^{n-1}j\\ &=\dfrac{n(n-1)}{2}\\ \end{array} $

The obvious next step is to get more precise bounds on $s(n)$.

I can show that $s(n) -(\frac{n^2}{2}-\frac{n}{3}) $ is bounded by writing

$\begin{array}\\ s(n) &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} (\sqrt{j^2+k}-j)\\ &=\sum_{j=1}^{n-1}j\sum_{k=0}^{2j} (\sqrt{1+k/j^2}-1)\\ \end{array} $

and using $1+x/2-x^2/8+x^3/16 \ge \sqrt{1+x} \ge 1+x/2-x^2/8 $.

(It turns out that using the upper bound $1+x/2 \ge \sqrt{1+x} $ is not enough, because a $\ln(n)$ term appears.)

The result I get is

$-(\frac1{6} +\frac{1}{24}\zeta(2)+\frac1{4}\zeta(3)+\frac{5}{48}\zeta(4)) \approx -0.648 \\ \le s(n) -(\frac{n^2}{2}-\frac{n}{3})\\ \le -\frac1{6}+\frac{5}{24}\zeta(2)+\frac1{16}\zeta(3) \approx 0.2511. $

Computation shows that the limit is about $-.2074 $.

So, my questions are:

Does the limit $\lim_{n \to \infty} s(n) -(\frac{n^2}{2}-\frac{n}{3}) $ exist?

If so, what is it?

I am sure that the limit exists, but I do not know what it might be.

from Hot Weekly Questions - Mathematics Stack Exchange