The sum itself:

$$ \frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+ \frac{1}{15}+ \frac{1}{39}... \le 1 $$

These are all sums of reciprocals of composites that can be closest to 1. Notice that the term after $\frac{1}{15}$ is $\frac{1}{39}$ rather than $\frac{1}{16}$, or any other reciprocal of a composite number between 15 and 39 for the reason that it would make the sum up to that point $\gt$ 1. That is a description of this sum from the most basic level.

Sum Parameters:

-If 1 is reached at an exact given point, the sum stops and thus making the sum finite.

-The sum must be $\le $ 1

-All terms must be the reciprocal of composite numbers.

-To find the next term in the list, you must find the largest reciprocal (that fits standards) that makes the sum $\le$ 1, and NOT the term that best fits the sum. For example, if you were to need to add $\frac{1}{x}$ and $\frac{1}{y}$ to make the sum exactly 1, but $\frac{1}{a}$ was the largest and still fit the rest of the parameters, then you would have to stick with with $\frac{1}{a}$. (Even if $\frac{1}{a}$ was a really ugly term to add at that point)


1) With these rules in place will the sum be a finite or infinite sum? If this is an infinite sum than the terms will always get closer and closer to 1, but never actually get there with a finite number of terms. With an finite sum, then at a quantifiable number of terms does the sum reach 1.

2)If the sum turns out to be infinite, than please tell me why.

3)If the sum turns out to be finite, list the terms. If the sum is ridiculously large, then prove that it is finite.

Thank you

  • $\begingroup$ I don't know this for sure (and at least some similar problems are unsolved), but it seems very likely that this sum would be finite whenever your target number was rational. $\endgroup$ – Micah Jul 6 '15 at 16:53
  • $\begingroup$ Its 1/6552. The number after the last one I wrote is that $\endgroup$ – AAron Jul 6 '15 at 16:56
  • $\begingroup$ Right, that's what happens when your target number is $1$. I'm talking about the more general problem when you're trying to get the sum to be any other rational number you like. $\endgroup$ – Micah Jul 6 '15 at 17:20
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    $\begingroup$ sister question $\endgroup$ – AAron Jul 6 '15 at 22:51

The sum is finite. The next term is the last: $1/6552$.

EDIT: Consider the more general problem with a rational target $T$.
I tried some random choices of target; for the target $105/37$ I was not able to produce a finite sum (after $204$ iterations the denominators were so big that Maple was having trouble with the primality testing). So I'm not convinced that the sum will always be finite.

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    $\begingroup$ Amazing. Now the question is why. Yeah, it just "comes out". But, I mean a deep why, if any. $\endgroup$ – ajotatxe Jul 6 '15 at 17:26
  • $\begingroup$ @ajotatxe: Time to post another question for $\frac{105}{37}$! Just a random guess: I think that as $m,n \to \infty$, the proportion of rational numbers with denominator in $[m..n]$ that results in an infinite sum would go to $1$. $\endgroup$ – user21820 Jul 7 '15 at 10:46
  • $\begingroup$ I can't finish $\frac{105}{37}$ either; Sage only gives me 150 terms, so apparently it's not nearly as good at primality testing as Maple. But I'm pretty sure it would terminate eventually. After getting over the initial hump where we need to use every available fraction, the numerators of the remainders start decreasing steadily. It's just that they have a long way to go, and meanwhile the denominators are increasing superexponentially... $\endgroup$ – Micah Jul 9 '15 at 16:52
  • $\begingroup$ ...yeah, unless there's a bug in my code, it looks like it terminates. The final denominator has $733833$ digits. $\endgroup$ – Micah Jul 10 '15 at 0:19
  • $\begingroup$ If you could prove that eventually the numerators will be strictly decreasing, that would be enough. $\endgroup$ – Robert Israel Jul 10 '15 at 0:39

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