Construct $\sqrt{3}$ using every natural number $n\in \mathbb{N}$ (1, 2, 3, 4...) exactly once and the operations addition ($+$) and division ($\div$).

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    $\begingroup$ I’m thinking this might be useful $\endgroup$
    – HTM
    Apr 10 '19 at 7:20
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    $\begingroup$ If it wasn't for your addition $(+)$ and division $(\div)$ rules, I would have used Ramanujan's identity, $$\sqrt{3} = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{\cdots}}}}}$$ $\endgroup$
    – Mr Pie
    Apr 10 '19 at 9:26
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    $\begingroup$ @user477343 And also, you use $1$ an infinite number of times. $\endgroup$
    – hexomino
    Apr 10 '19 at 9:35
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    $\begingroup$ @hexomino oh yes, true; I forgot about using every number once. Thanks for that! (You can have an upvote for that :P) $\endgroup$
    – Mr Pie
    Apr 10 '19 at 10:01
  • $\begingroup$ Does the question ask that each natural number (besides 0) is used at least once or exactly once? The original wording was ambiguous, but I assumed exactly once because at least once would be relatively straightforward. But I think the answer posted by the OP uses integers multiple times. $\endgroup$
    – noedne
    Apr 11 '19 at 2:51

It starts like



use the natural numbers in their natural order, keeping track of the current sum of fractions you have produced so far. If $n$ is the next number to be used, try to add $\frac{n}{n+1}$. If that increases the sum above $\sqrt{3}$, append the next number ($n+2$) to the denominator, and go on with the same procedure with $n+3$, then $n+4$ if necessary, and so on. Sooner or later the sum will get below $\sqrt{3}$. It is then time to start a new fraction.

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    $\begingroup$ Is it proven (e.g. by limit to infinity) that the result will be $\sqrt 3$? Can I start with arbitrary constant in the beginning? $\endgroup$
    – athin
    Apr 10 '19 at 6:22
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    $\begingroup$ The limit can be any positive constant, as the series of $\frac{2n+1}{2n+2}$ is divergent, so summing those terms we can get arbitrarily large values, and then use the 'rest' for finetuning the limit in infinity. In this case the limit of $\sqrt{3}$ is reached as a limit exactly by our definition which involved a decision based on the partial sums being higher or lower than $\sqrt{3}$. Had we used $\pi^\pi$, that'd have been the limit. $\endgroup$
    – elias
    Apr 10 '19 at 6:46
  • $\begingroup$ While it is clear that the series will not exceed $\sqrt{3}$, I don't think it is immediately obvious that it reaches that limit. Could it not be the case that it converges to a smaller number? That it doesn't do that depends not only on the fact that $\frac{2n+1}{2n+2}$ diverges, but also that the series with k numbers in the denominator of every term is a divergent series for all k>1. $\endgroup$ Apr 10 '19 at 6:49
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    $\begingroup$ You're right, but I'd argue that it's the same 'level' of being obvious: $\frac{(k+1)n+1}{(k+1)n+2+\dots+(k+1)n+k+1}<\frac{(k+1)(n+1)+1}{(k+1)(n+1)+2+\dots+(k+1)(n+1)+k+1}$, and you can see this even without computing these: take the elements of the left and right hand side pairwise. The one that grows the most (in ratio) is the one in the numerator (as that was the smallest one, and they all have the same amount added). $\endgroup$
    – elias
    Apr 10 '19 at 7:01
  • $\begingroup$ Great answer! @JaapScherphuis I think its easy to see that it reaches the limit because $ \lim_{n\to \infty}{\frac{n}{\sum_{i=0}^k n+i}-\frac{n}{\sum_{i=0}^{k+1} n+i}}=0$ and $\lim_{n\to \infty}{\frac{n}{n+1}}=1 $. For large n, if you go over $\sqrt(3)$ you're essentially stuck subtracting very small values until you're under $\sqrt(3)$, then add 1 and back again. The series can't converge to any other constant $\endgroup$
    – Artemmm
    Apr 10 '19 at 9:42

Here's a formula which I believe contains each number once. Multiplication isn't converted to division for readability's sake.

$\sqrt{3}$ = $$1 *\frac{3+5}{2+4}* \prod_{n=0}^\infty \frac{\Big[\prod_{k=1}^{13}(6(13n+k)+2)(6(13n+k)+4)\Big]^a}{ \Big[\prod_{k=1}^{13}(6(13n+k)+3)\Big]^b \prod_{k=1}^{13}(6(13n+k)+1)(6(13n+k)+5)}$$


$$ a=\frac{6(13n)+6(13n+2)+6(13n+4)}{6(13n+1)+6(13n+3)}*\frac{6(13n+5)+6(13n+7)}{6(13n+6)}$$ and $$ b=\frac{6(13n+8)+6(13n+9)+6(13n+11)+6(13n+12)}{6(13n+10)}$$

I derived it by combining the Wallis product for pi with this infinite product for sine evaluated at x=$\frac{\pi}{3}$:

$$ \sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right) $$

When all terms of the form $6(13n+k)+c$ are written out the product should contain each natural number once. $a=3$ and $b=4$ for all n


$\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$ so $\sqrt{3}=2*\sin{\frac{\pi}{3}}$

I can expand this using the above formula for sine:

$$\sin{\frac{\pi}{3}} = \frac{\pi}{3}\prod_{n=1}^\infty \left(1-\frac{\frac{\pi^2}{3^2}}{n^2\pi^2}\right)=\frac{\pi}{3}\prod_{n=1}^\infty \left(\frac{9n^2-1}{9n^2}\right)=\frac{\pi}{3}\prod_{n=1}^\infty \left(\frac{(3n-1)(3n+1)}{3n*3n}\right) $$

Now, the Wallis product states that

$$\prod_{k=1}^{\infty} \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2} $$

which means

$$\frac{\pi}{3}=\frac{2}{3} \cdot \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots $$

Plugging this into the infinite product for $\sin{\frac{\pi}{3}}$ we get

$$\sin{\frac{\pi}{3}}=\frac{2}{3} \cdot \frac{A}{B} $$

where the placement of every natural number in $A$ or $B$ depends on its mod 6 residue. Numbers divisible by 6 cancel out and I used them to construct a=3 and b=4 above. The term at the start is just $\frac{3+5}{2+4}=\frac{4}{3}=2*\frac{2}{3}$

  • $\begingroup$ It must use each number exactly once, I'm afraid. Beautiful formula, nevertheless! :) $\endgroup$
    – Mr Pie
    Apr 11 '19 at 3:58
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    $\begingroup$ @user477343 Each number occurs in the expression $\bf{exactly} $ one time $\endgroup$
    – Artemmm
    Apr 11 '19 at 4:16
  • $\begingroup$ How do you convert from multiplication to division without using 1s? $\endgroup$
    – noedne
    Apr 11 '19 at 4:27
  • $\begingroup$ What about the number $6$ for example? $$\sqrt{3}=1\times\frac{3+5}{2+4}\times \prod_{n=0}^\infty \frac{\bigg[\prod\limits_{k=1}^{13}(6(13n+k)+2)(\color{red}{6}(13n+k)+4)\bigg]^a}{ \bigg[\prod\limits_{k=1}^{13}(\color{red}{6}(13n+k)+3)\bigg]^b \prod\limits_{k=1}^{13}(\color{red}{6}(13n+k)+1)(\color{red}{6}(13n+k)+5)}$$ You have four extra $6$'s (not including the values of $a$ and $b$). $\endgroup$
    – Mr Pie
    Apr 11 '19 at 4:27
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    $\begingroup$ @noedne The conversion is just rewriting $\frac{a*b}{c}$ as $\frac{a}{\frac{c}{a}}$ $\endgroup$
    – Artemmm
    Apr 11 '19 at 4:31

You can do this


It is constructed like this:

Start with 2/1. Then, for every n = 2, 3, ..., if the partial product is below your target of sqrt(3) then multiply it by (2n)/(2n-1) else multiply it by (2n-1)/(2n).
Every new factor makes the product move towards the target, sometimes overshooting, but never more than by a ratio of (2n)/(2n-1). Since that ratio converges to 1, the product converges to the target.

It should work for every positive target.

  • $\begingroup$ I guess you mean that the ratio converges to 1 and thus the ratio between the product and the target also converges to 1... this also requires proving that prod(evens)/prod(odds) -> Infinity (i.e.., that you cannot "undershoot forever"), but that does hold. $\endgroup$
    – wimi
    Jun 17 at 12:16
  • $\begingroup$ Indeed, the ratio converges to 1, not 0. I corrected that. Regarding the limit of prod((2n)/(2n-1)), you can show that it is larger than sum(1/(2n-1)) and that diverges. $\endgroup$
    – Florian F
    Jun 17 at 14:00

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