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The following is not a homework, just curiosity.

Consider the integers grouped by consecutive pairs : $(1,2)$, $(3,4)$, ...

What is the limit of the "switching fractions" where we alternatively use the largest number in a pair either upward or downward :

$$\frac 12, \frac{1\cdot 4}{2 \cdot 3}, \frac{1\cdot 4\cdot 5}{2\cdot 3\cdot 6}, \frac{1\cdot 4\cdot 5\cdot 8}{2\cdot 3\cdot 6\cdot 7},\ldots?$$ A proof as elementary as possible would be nice, if not it could use standard results on prime distribution.

Also, was it considered before? Any reference welcomed.

Edit Numerically we have: $0.5, 0.6666... , 0.5555... , 0.6349206..,0.5714286..,0.6233766...$ More terms would certainly help.

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    $\begingroup$ Did you try computing a few terms and see if the limit approaches anything? See how to ask a good question. $\endgroup$
    – Pedro
    Jan 10, 2023 at 12:53
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    $\begingroup$ The limit seems to be $$ \sqrt \pi \frac{{\Gamma \big( {\frac{3}{4}} \big)}}{{\Gamma\! \left( {\frac{1}{4}} \right)}} = 0.5990701 \ldots $$ You can try using the Weierstrass product. $\endgroup$
    – Gary
    Jan 10, 2023 at 13:00
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    $\begingroup$ @Gary that's just beautiful $\endgroup$ Jan 10, 2023 at 13:00
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    $\begingroup$ Taking the logarithm of your sequence, it is easier to see that it actually converges. $\endgroup$
    – Arthur
    Jan 10, 2023 at 13:05
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    $\begingroup$ It is essentially the $5$th result here : en.wikipedia.org/wiki/Lemniscate_constant#Series (Take the reciprocal and divide by $2$.) $\endgroup$
    – Gary
    Jan 10, 2023 at 13:06

1 Answer 1

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A more simpler, yet heavier way to look at this is,

A more common series is given by $\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\ldots=\prod\limits^{n}_{k=1}{\dfrac{2k-1}{2k}}=\dfrac{\left(n-\frac{1}{2}\right)!}{n!\cdot\left(\frac{-1}{2}\right)!}$.

Your problem is slight altered with every alternate term inversed, $\dfrac{1}{2}\cdot\left(\dfrac{3}{4}\right)^{-1}\cdot\dfrac{5}{6}\cdot\left(\dfrac{7}{8}\right)^{-1}\cdot\ldots$

In fact we factor with respect to the inverse power, $$\left(\frac{1}{2}\cdot \frac{5}{6}\cdot\ldots\right)\cdot\left(\frac{3}{4}\cdot\frac{7}{8}\cdot\ldots\right)^{-1}$$

$$=\left(\prod^{n}_{k=1}{\frac{4k-3}{4k-2}}\right)\cdot\left(\prod^{n}_{k=1}{\frac{4k-1}{4k}}\right)^{-1}$$ $$=\left(\frac{\left(n-\frac{1}{4}\right)!}{n!\cdot \left(\frac{-1}{4}\right)!}\right)^{-1}\cdot\frac{\left(n-\frac{3}{4}\right)!\cdot\left(\frac{-1}{2}\right)!}{\left(n-\frac{1}{2}\right)!\cdot\left(\frac{-3}{4}\right)!}$$

Funnily enough, the limit does converge for the above expression ($n\to\infty$); $$\frac{\left(\frac{-1}{4}\right)!\cdot \left(\frac{-1}{2}\right)!}{\left(\frac{-3}{4}\right)!}\approx 0.599195 \text{ as mentioned in answers above}$$

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    $\begingroup$ Nicely elementary! It also suggests variants like $\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\left(\dfrac{5}{6}\right)^{-1}\cdot\dfrac{7}{8}\cdot\dfrac{9}{10}\cdot\left(\dfrac{11}{12}\right)^{-1}\cdot\ldots$ $\endgroup$ Jan 10, 2023 at 13:56

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