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

Question

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.

Answer 1

It's annoying to realize that taking a different path, as suggested in the comments, would have provided a direct solution.

I make use of the result of Marko Riedel linked to by EDX and Gary that $\sum_{k=1}^n\sqrt{k} \sim \frac23 n^{3/2}+\frac12\sqrt {n} +\zeta \left( -\frac12 \right) + {\frac {1}{24\sqrt {n}}} -{\frac {1}{1920{n}^{5/2}}}\,+{\frac {1}{9216{n}^{9/2}}} +\cdots $.

This implies that

$\sum_{k=1}^{n^2}\sqrt{k} \sim \frac23 n^3+\frac{n}{2} +\zeta \left( -\frac12 \right) + {\frac {1}{24n}} -{\frac {1}{1920n^5}}+\frac {1}{9216n^9} +\cdots $

so that

$\sum_{k=1}^{n^2-1}\sqrt{k} \sim \frac23 n^3-\frac{n}{2} +\zeta \left( -\frac12 \right) + {\frac {1}{24n}} -{\frac {1}{1920n^5}}+\frac {1}{9216n^9} +\cdots $

I find it amusing that this result was in response to a question of mine.

$\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}-\sum_{j=1}^{n-1}\sum_{k=0}^{2j} j\\ &=s_1(n)-s_2(n)\\ s_1(n) &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} \sqrt{j^2+k}\\ &=\sum_{j=1}^{n-1}\sum_{k=j^2}^{(j+1)^2-1} \sqrt{k}\\ &=\sum_{k=1}^{n^2-1} \sqrt{k}\\ &=\dfrac23 n^3-\dfrac{n}{2} +\zeta \left( -\dfrac12 \right) + {\dfrac {1}{24n}} -{\dfrac {1}{1920n^5}}+\dfrac {1}{9216n^9} +\cdots\\ s_2(n) &=\sum_{j=1}^{n-1}\sum_{k=0}^{2j} j\\ &=\sum_{j=1}^{n-1}j(2j+1)\\ &=2\sum_{j=1}^{n-1}j^2+\sum_{j=1}^{n-1}j\\ &=2\dfrac{(n-1)n(2n-1)}{6}+\dfrac{n(n-1)}{2}\\ &=\dfrac{2n^3-3n^2+n}{3}+\dfrac{n(n-1)}{2}\\ &=\dfrac{4n^3-6n^2+2n+3n^2-3n}{6}\\ &=\dfrac{4n^3-3n^2-n}{6}\\ &=\dfrac{2n^3}{3}-\dfrac{n^2}{2}-\dfrac{n}{6}\\ \text{so}\\ s(n) &=s_1(n)-s_2(n)\\ &=\dfrac{n^2}{2}-\dfrac{n}{3} +\zeta \left( -\dfrac12 \right) + {\dfrac {1}{24n}} -{\dfrac {1}{1920n^5}}+\dfrac {1}{9216n^9} +\cdots\\ \end{array} $

This agrees with Gary's comment.