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Let's call the following numbers than can be produced by playing with plus and minus: $$H_n'=\pm\frac{1}{1}\pm\frac{1}{2}\pm\frac{1}{3}\pm\cdots\pm\frac{1}{n}$$
"Harmonic kids" of $H_n$.
We have a free choice of plus and minus for every term,so there are $2^{n}$ "Harmonic kids".

Is it possible to find out which one of them has the smallest absolute value?
(And even better:it's value!!!)
Thanks in advance!

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  • $\begingroup$ I don't have any method, but I suspect either $$1-\frac{1}{2}+\frac{1}{3}-\cdots$$ or,$$1\underbrace{-\frac{1}{2}-\frac{1}{3}}\,\underbrace{+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}\,\underbrace{-\frac{1}{8}\cdots-\frac{1}{15}}\,+\cdots$$ $\endgroup$ – Guy Apr 6 '14 at 16:08
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    $\begingroup$ Some remarks: You can never get exactly $0,$ since $$n! H_n' = \pm 2\cdots n \pm 1\cdot 3 \cdots n \pm 1\cdot 2\cdot 4 \cdots n \pm \cdots \pm 1\cdot 2 \cdot 3 \cdots (n-1) $$ is not zero modulo $p,$ where $p$ is some prime $n/2 +1 < p < n$ (whose existence is given by Bertrand's postulate). The crude method of adding the first $k$ terms of $H_n'$ and subtracting the rest has the best result when you pick $k\approx \sqrt{n/e^{\gamma}}$ where $\gamma$ is the Euler-Mascheroni constant, in which case you get very close to $1/k.$ $\endgroup$ – Ragib Zaman Apr 6 '14 at 17:08
  • $\begingroup$ @Sabyasachi i do not think that this is the best result we can get but it certainly is a good idea.... $\endgroup$ – Konstantinos Gaitanas Apr 6 '14 at 19:41
  • $\begingroup$ @RagibZaman interesting. I just made the alternating $+$ and $-$ with the geometric series in mind(i dont really know why), but $\gamma$ is more fitting here, indeed. $\endgroup$ – Guy Apr 6 '14 at 20:40
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This is A232111(n)/A232112(n):

1/1, 1/2, 1/6, 1/12, 7/60, 1/20, 11/420, 13/840, 11/2520, 11/2520, 23/27720, 23/27720, 607/360360, 251/360360, 251/360360, 25/144144, 97/12252240, ...

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If it helps, these are the answers for the first few cases. "1" stands for adding and "-1" for subtracting. For instance, $[1,1,-1]$ would mean the number ${1\over 1}+{1\over 2}-{1\over 3}$. The best choices are $$[-1, 1], [-1, 1, 1], [-1, 1, 1, 1], [1, -1, -1, -1, 1], [-1, 1, 1, 1, -1, 1], [1, -1, -1, -1, -1, 1, 1], [-1, 1, 1, 1, -1, -1, 1, 1], [-1, 1, 1, -1, 1, -1, 1, 1, 1], [1, -1, -1, -1, 1, 1, -1, -1, -1, 1], [-1, 1, 1, 1, -1, -1, -1, 1, 1, 1, 1], [1, -1, -1, -1, 1, -1, 1, 1, -1, -1, -1, 1], [-1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, 1, 1], [-1, 1, 1, 1, 1, -1, -1, -1, 1, -1, -1, 1, 1, 1]$$ I used a program to find them, and I'm pretty sure they are accurate. The sums are $$0.5, 0.16666666666666669, 0.08333333333333331, 0.1166666666666667, 0.04999999999999996, 0.02619047619047618, 0.015476190476190477, 0.00436507936507935, 0.004365079365079377, 0.000829725829725847, 0.0008297258297258053, 0.0016844266844266292, 0.0006965256965257016$$

I have added a pictorial representation below. That may help to find a pattern.

-+
-++
-+++
+---+
-+++-+
+----++
-+++--++
-++-+-+++
+---++---+
-+++---++++
+---+-++---+
-+-++-++++-++
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  • $\begingroup$ I can add more if that's helpful... $\endgroup$ – user138335 Apr 6 '14 at 16:24
  • $\begingroup$ This is useful, thanks. It would also be helpful if you could put the corresponding value of the sum for each of these. $\endgroup$ – Ragib Zaman Apr 6 '14 at 16:51
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    $\begingroup$ Since we're talking about absolute value, +---+ is identical to -+++-. Having all the series start with the same sign would make comparison easier. $\endgroup$ – Zachiel Apr 6 '14 at 22:20

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