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 $C=\zeta(-\frac 12)\approx-0.207886224977...$.

Solution 1:

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.

Solution 2:

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.