How close can $\sum_{k=1}^n \sqrt{k}$ be to an integer?

How close can
$S(n) = \sum_{k=1}^n \sqrt{k}$
be to an integer?
Is there some $f(n)$ such that,
if $I(x)$ is the closest integer to $x$,
then $|S(n)-I(S(n))|\ge f(n)$
(such as $1/n^2$, $e^{-n}$, …).

This question was inspired by the recently proposed and answered question of
“prove that $\sum_{k=1}^n \sqrt{k}$
is never an integer/”.
The question is here: Is $\sqrt1+\sqrt2+\dots+\sqrt n$ ever an integer?

The Euler-Maclaurin estimate for $S(n)$
might be useful.
According to an answer here,
(the link is “How to calculate the asymptotic expansion of $\sum \sqrt{k}$?”)
$$S(n) = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2})$$
$C=\zeta(-\frac 12)\approx-0.207886224977…$.

Solutions Collecting From Web of "How close can $\sum_{k=1}^n \sqrt{k}$ be to an integer?"

Thanks Marty for a fascinating question. We can get the entire
asymptotic expansion quite easily using Mellin transforms.

Start with the telescoping sum
$$ S(x) = \sum_{k\ge 1} \left(\sqrt{k}-\sqrt{x+k}\right)$$
which has the property that $$ S(n) = \sum_{q=1}^n \sqrt{q}$$
so that $S(n)$ is the value we are looking for.

Now re-write the inner term so that we can see the harmonics:
$$ \sqrt{k}-\sqrt{x+k} = \sqrt{k}\left(1-\sqrt{x/k+1}\right).$$

Now recall that
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x); s\right)=
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right)g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \sqrt{k}, \quad \mu_k = \frac{1}{k}
\quad \text{and} \quad g(x) = 1-\sqrt{1+x}.$$

It follows that $$ \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} =
\sum_{k\ge 1} \sqrt{k} \times k^s =\zeta(-1/2-s).$$

Furthermore we have
$$\mathfrak{M}(g(x); s) = \frac{1}{2\sqrt{\pi}} \Gamma(-1/2-s)\Gamma(s).$$

Now this transform has fundamental strip $\langle -1, -1/2 \rangle$ while the zeta function term has $-s-1/2 > 1$ or $s < -3/2.$ These two are disjoint. Therefore we need to modify $g(x)$ by canceling the next term in the power series of $-\sqrt{1+x},$
which gives $$g(x) = 1 + \frac{1}{2} x – \sqrt{x+1},$$ with fundamental strip
$\langle -2, -1 \rangle,$ and the transform of $g(x)$ being the same. This strip is perfect as the half-plane of convergence of the zeta function term starts right in the middle of it, extending to the left.

It is important to note that we have now added $$\sum_{k\ge 1} \frac{1}{2}\sqrt{k} \frac{x}{k} = \frac{1}{2} x \sum_{k\ge 1} \frac{1}{\sqrt{k}} = \frac{1}{2} x \zeta(1/2)$$ to our sum, which we will have to subtract out at the end.

The conclusion is that the Mellin transform $T(s)$ of $S(x)$ is given by
$$T(s) = \frac{1}{2\sqrt{\pi}} \Gamma(-1/2-s)\Gamma(s) \zeta(-1/2-s).$$

Now apply Mellin inversion, shifting the integral
$$\frac{1}{2\pi i}\int_{-7/4-i\infty}^{-7/4+i\infty} T(s)/x^s ds$$
to the right to obtain an expansion at infinity.

We obtain that
$$\operatorname{Res}(T(s)/x^s; s=-3/2) = -\frac{2}{3} x^{3/2},$$
$$\operatorname{Res}(T(s)/x^s; s=-1) = -\frac{1}{2} \zeta(1/2) x,$$
(this residue does not contribute being cancelled by the term that we introduced to shift the fundamental strip of $g(x)$)
$$\operatorname{Res}(T(s)/x^s; s=-1/2) = -\frac{1}{2} x^{1/2},$$
$$\operatorname{Res}(T(s)/x^s; s=0) = -\zeta(-1/2),$$
$$\operatorname{Res}(T(s)/x^s; s=1/2) = -\frac{1}{24} x^{-1/2}.$$
The remaining residues have the form
$$\operatorname{Res}(T(s)/x^s; s=2q+1/2) =
Here we use $q\ge 1.$ The reader may wish to simplify these.

This yields the asymptotic expansion
$$S(n) \sim
2/3\,{n}^{3/2}+1/2\,\sqrt {n}+\zeta \left( -1/2 \right) +
1/24\,{\frac {1}{\sqrt {n}}}
-{\frac {1}{1920}}\,{n}^{-5/2}+{\frac {1}{9216}}\,{n}^{-9/2} +\cdots$$

This is as it ought to be and here Mellin transforms really shine. Mellin-Perron and Wiener-Ikehara only give the first few terms while Euler-MacLaurin fails to produce the constant. The following MSE link points to a calculation in a very similar spirit.

Suppose we are given a small $\epsilon\gt 0$. We show that we can choose $n$ such that $\sum_1^n\sqrt{k}$ is within $\epsilon$ of an integer. Relatively simple estimates are used.

For take $N\gt 1/\epsilon$. Then the numbers
$$\sqrt{N^4+1},\quad \sqrt{N^4+2},\quad\text{and so on up to}\quad \sqrt{N^4+2N}
are greater than $N^2$ but within $\epsilon$ of $N^2$. Thus each of them is “nearly” an integer. To see this, note that $\left(N^2+\frac{1}{N}\right)^2 \gt N^4 +2N$.

Moreover, these numbers have fractional parts that add up to more than $1$. This is fairly straightforward, since the smallest fractional part is approximately $\frac{1}{2N}$.

So however far $\sum_1^{N^4} \sqrt{k}\,\,$ may be from an integer, one of the sums to $n=N^4+i$, where $1\le i\le 2N$, must come within $\epsilon$ of an integer.

Remark: It may be interesting to ask how much better one can do than $M\approx \frac{1}{\epsilon^4}$ to be sure that there is an $n\le M$ such that $\sum_1^n\sqrt{k}$ is within $\epsilon$ of an integer. Presumably much better! That is where more sophisticated ideas such as Euler-Maclaurin may be useful.

The form for the asymptotic expansion suggests examining values of $n$ near $36m^2$, where $m\in\mathbb{Z}^+$.
For $a\in\mathbb{Z}$ and $|a|\ll m^2$ we find
S(36m^2+a) = 144m^3+(6a+3)m+\frac{a(a+1)}{24m}+\zeta(-1/2) + O(1/m).\tag{1}
Thus, if $m$ is large and
m = \left[-\frac{a(a+1)}{24\zeta(-1/2)}\right],\tag{2}
where $[\;]$ is the nearest integer function,
the sum itself should be near an integer.
The expansion (1) is consistent, since for this choice of $m$ we must have $|a|\sim\sqrt{m}\ll m^2$.

The error introduced by (2) is $O(1/m)$.
Thus, the sum for this choice of $n$ will be within $O(1/m)\sim O(1/n^{1/2})$ of an integer.

Below we give the distance to the nearest integer to the seventh decimal place for some values of $n$.
a & n & |S(n)-[S(n)]| \\ \hline
2 & 38 & 0.0462347 \\
4 & 580 & 0.0019127 \\
8 & 7064 & 0.0068525 \\
16 & 108916 & 0.0017046 \\
32 & 1618016 & 0.0003070 \\
64 & 25040080 & 0.0000443 \\
128 & 394419728 & 0.0000292

To complete this calculation we need to show how to compute
$$g^*(s) = \mathfrak{M}(\sqrt{x+1}; s).$$
This is
$$\int_0^\infty \sqrt{x+1} x^{s-1} dx.$$
Now put $x+1 = 1/t$ to get
$$ g^*(s) = \int_1^0 \frac{1}{\sqrt{t}} \frac{(1-t)^{s-1}}{t^{s-1}}
\left(-\frac{1}{t^2}\right) dt \\ =
\int_0^1 t^{-1/2-s+1-2} (1-t)^{s-1} dt =
\int_0^1 t^{-s-3/2} (1-t)^{s-1} dt.
This last integral is a beta function term and equal to
$$B(-s-1/2, s) = \frac{\Gamma(-s-1/2)\Gamma(s)}{\Gamma(-1/2)} =
This was to be shown.