# $S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$ is almost an integer. Find $m$

For an integers $$m$$ and $$n$$, $$1 , we need to find the best $$m$$ so that $$S=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots+\sqrt{m}$$ is almost an integer.

• Example: when $$n=40$$, then the best value of $$m$$ is $$38$$, where $$S$$ become $$159.046\dots$$

I used Microsoft Excel and observes the sums, term-by-term, of the series, which is not a sufficient way for large values of $$n$$.

Using mathematical equations and formulae, not programs like Excel,

How to find the best value of $$m$$ if we are given larger values of $$n$$?

• Define "almost".
– lhf
Jul 9, 2019 at 17:59
• Estimating the tail of the sum by an integral could work. Jul 9, 2019 at 17:59
• @lhf, thanks for asking me to define, I should define it without your question. I mean the best value of $m$ so that $|[S]-S|$ attains its minimum, where [ ] denotes the round function. Jul 9, 2019 at 18:03
• Do you know the asymptotic of $f(m) = \sum_{n=1}^m \sqrt{n}$ ? Do you know how to improve the error term (the first steps of the Euler-Maclaurin summation formula) ? Jul 9, 2019 at 18:06
• Jul 9, 2019 at 18:08

Let $$F(m) \equiv \left| \sum_{n=1}^{m}\sqrt{n} - \left[ \sum_{n=1}^{m}\sqrt{n} \right] \right|$$ where $$[x]$$ denotes the nearest integer to $$x$$. A small value of $$F(m)$$ tells you that the sum of the first $$m$$ square roots is near to an integer.

What you are really looking for is the sequence of $$m_i$$ where $$m_i$$ is monotonic increasing with $$i$$ and for each $$m_i$$ $$k < m_i \Longrightarrow F(k) > F(m_i)$$ That is, the sequence of "closest to an integer yet" values of $$m$$.

The easy way to see the start of this sequence is to note that the asymptotic form of $$\sum_1^m\sqrt{n}$$ is $$\sum_1^m\sqrt{n}\approx \frac1{\sqrt{n}}\left(\frac{2n^2}{3}+\frac{n}2 +\zeta(-\frac12) \sqrt{n}+\frac1{24}-\frac1{1920 n^2}+ \frac1{9216 n^4}-\frac{11}{163840 n^6}+\frac{65}{786432 n^8}\right)$$ Then in Mathematica you gan define $$F[m]$$ as that expression, $$G[m]$$ as Abs$$[F[m]-$$Round$$[F[m]]$$ and do a series of DiscretePlot of $$G[m]$$ to see where you get new minima.

When you do this for up to $$m=10^6$$ you find the sequence for $$m_i$$ is $$\{ 3, 13, 22, 33, 38, 41, 54, 156, 761, 10869, 41085, 142625, 224015, 898612\ldots\}$$ So for example, when $$n=500$$ the best value of $$m$$ will be $$156$$, at which point $$\sum_1^{156}\sqrt{n}\approx 1305.0000314264$$ The sequence $$m_i$$ given above is not in OEIS.

The next number in the sequence is $$2750788$$ and $$\sum_1^{2750788}\sqrt{n}\approx 3041547064.000000030776$$ That nearness ($$3\cdot 10^{-8}$$) is a significant improvement over its predecessor, which is off by a bit more than one part in a million.

• I thought that this problem was beyond the reach of mortals. Jul 9, 2019 at 20:43
• Can we use $(m+a)^{1/2} = m^{1/2}+ \frac{a}2 m^{-1/2}+O(a^2 m^{-3/2})$ to find the $m$ without computing the function ? Jul 9, 2019 at 22:47
• You really should submit this to OEIS. Jul 9, 2019 at 23:01