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 $\sum \sqrt{k}$?")
$$S(n) =  \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2})$$
where
$C=\zeta(-\frac 12)\approx-0.207886224977...$.
 A: Contrary to the numbers obtained by the clever approach of @user26872, The sequence of "best values up to this point" for numbers $m$ such that $\sum_1^m\sqrt{n}$ is closer to an integer than it is for any smaller $n$, is
$$
\{ 3, 13, 22, 33, 38, 41, 54, 156, 761, 10869, 41085, 142625, 224015, 898612, 2750788 \ldots\}
$$
For example, 
$\sum_1^{156}\sqrt{n} \approx 1305.000314$ and 
$\sum_1^{10869}\sqrt{n} \approx 755479.999989$ and 
$\sum_1^{2750788}\sqrt{n}$ is only $3.0776 \cdot 10^{-8}$ more than an integer.
Of these best values, only $38$ is on the list given in the other answer.
The observation is that you can get better approaches to an integer by allowing the leading terms to stray off of integer values, in clever ways that almost cancel the lower order terms.
A: 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.
A: 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
$$B(-s-1/2, s) = \frac{\Gamma(-s-1/2)\Gamma(s)}{\Gamma(-1/2)} =
-\frac{\Gamma(-s-1/2)\Gamma(s)}{2\sqrt{\pi}}.$$
This was to be shown.
A: 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
$$\operatorname{Res}(T(s)/x^s; s=2q+1/2) = 
\frac{1}{2\sqrt{\pi}}\Gamma(2q+1/2)\zeta(-2q-1)\frac{x^{-2q-1/2}}{(2q+1)!}.$$
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.
A: 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 
$$\begin{equation*}
S(36m^2+a) = 144m^3+(6a+3)m+\frac{a(a+1)}{24m}+\zeta(-1/2) + O(1/m).\tag{1}
\end{equation*}$$
Thus, if $m$ is large and 
$$\begin{equation*}
m = \left[-\frac{a(a+1)}{24\zeta(-1/2)}\right],\tag{2}
\end{equation*}$$
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 $n$.
$$\begin{array}{lll}
a & n & |S(n)-[S(n)]| \\ \hline
 2 & 38 &           0.0462347 \\
 4 & 580 &          0.0019127 \\
 8 & 7064 &         0.0068525 \\
 16 & 108916 &      0.0017046 \\
 32 & 1618016 &     0.0003070 \\
 64 & 25040080 &    0.0000443 \\
 128 & 394419728 &  0.0000292
\end{array}$$
