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I thought it was easy to show that $\sqrt {2 \sqrt {3 \sqrt {4 \ldots}}}$ is irrational, but found a gap in my proof. Simple finite approximations show the denominator cannot be small, though, strongly suggesting irrationality. However, can it be shown whether this number is algebraic or transcendental? My hunch is that it's transcendental but I have no idea on how to start such a proof (especially since I can't fill my gap to prove irrationality).

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    $\begingroup$ This is related to one of Ramanujan's identities for nested radicals - see en.wikipedia.org/wiki/Nested_radical - someone will post a neat version of this as an answer I am sure. $\endgroup$ – Mark Bennet Aug 15 '14 at 15:44
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    $\begingroup$ I am pretty sure it's an open problem. OEIS doesn't mention anything about transcendence. $\endgroup$ – Balarka Sen Aug 15 '14 at 15:46
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    $\begingroup$ To make something of the question, you might want to post a proof that the number is irrational. $\endgroup$ – Did Sep 26 '14 at 17:33
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    $\begingroup$ It can be written as the infinite product: $2^{\frac{1}{2}}*3^{\frac{1}{4}}*4^{\frac{1}{8}}*5^{\frac{1}{16}}...$, all of which are algebraic, but I don't know if that helps establish transcendence or not. $\endgroup$ – Ari Apr 15 '15 at 21:24
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    $\begingroup$ 12 answers posted, 11 of them deleted. Wonder whether that's a record number of deleted answers. $\endgroup$ – Gerry Myerson Jun 21 '15 at 23:32
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This is not an answer, just a list of papers concerned with this constant; it's too long for a comment. Going by the reviews of the papers, they deal with efficient methods for calculating the constant, and not (directly) with questions of irrationality and/or transcendence. The first paper on the list is too new to have been reviewed at this point.

MR3349435
Lu, Dawei; Song, Zexi; Some new continued fraction estimates of the Somos' quadratic recurrence constant. J. Number Theory 155 (2015), 36–45.

MR3019753
Chen, Chao-Ping New asymptotic expansions related to Somos' quadratic recurrence constant.
C. R. Math. Acad. Sci. Paris 351 (2013), no. 1-2, 9–12.

MR2825112 (2012h:11181)
Hirschhorn, Michael D. A note on Somos' quadratic recurrence constant.
J. Number Theory 131 (2011), no. 11, 2061–2063.

MR2809034 (2012e:05038)
Nemes, Gergő On the coefficients of an asymptotic expansion related to Somos' quadratic recurrence constant.
Appl. Anal. Discrete Math. 5 (2011), no. 1, 60–66.

MR2684487 (2011i:11179)
Mortici, Cristinel Estimating the Somos' quadratic recurrence constant.
J. Number Theory 130 (2010), no. 12, 2650–2657.

MR2319662 (2008f:40013)
Sondow, Jonathan; Hadjicostas, Petros The generalized-Euler-constant function $\gamma(z)$ and a generalization of Somos's quadratic recurrence constant.
J. Math. Anal. Appl. 332 (2007), no. 1, 292–314.

MR2262724 (2008b:11081)
Hessami Pilehrood, Khodabakhsh; Hessami Pilehrood, Tatiana Arithmetical properties of some series with logarithmic coefficients.
Math. Z. 255 (2007), no. 1, 117–131.

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  • $\begingroup$ Thanks, if any of these papers either settles my question or shows me how to do so myself, I'll post another bounty and give it to your answer. I really appreciate your efforts finding all these papers. I had no idea about the term "Somo's quadratic recurrence constant" for example. $\endgroup$ – user2566092 Jun 26 '15 at 17:27
  • $\begingroup$ There was a reference to Somos' quasdratic recurrence constant in the comment by @PeterWoolfitt from 15 August 2014. My understanding of the way the bounty system works is that if the person offering the bounty doesn't award it, then half of it disappears but half of it goes to the person with the highest voted answer. I'm sure the details on that are available somewhere on the site. $\endgroup$ – Gerry Myerson Jun 27 '15 at 5:58
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There exists a finite difference formulation of this problem. This doesn't solve your problem, but it may be at least a bit enlightening, Consider $f(x)$ such that

$$ f(x) = x\sqrt{f(x+1)} $$

Then it follows

$$ f(x) = x \sqrt{(x+1) \sqrt{(x+2) \sqrt{... }}} $$

And $f(1)$ is the value you wish to evaluate.

Another way to write that is

$$ \frac{f(x)^2}{x^2} = f(x+1) $$

Let $f(x) = 2^{g(x)} $

$$ \frac{2^{2g(x)}}{x^2} = 2^{g(x+1} $$

Taking the logarithm of both sides base-2

$$ 2g(x) - 2\log_2{x} = g(x+1) $$

We can now evaluate, this as a run of the mill finite difference equation in G

$$ D_{1,x}g - g = -2\log_2{x} $$

(... lots of work that you can comment and request if you want ...)

$$ 2^{1-x}g = -D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$

So it follows that

$$ g = -2^{x-1}D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$

And that

$$ f = 2^{ -2^{x-1}D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right]} $$

Note that $f(0) = 0$ and therefore the expression

$$ D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$

must tend to positive infinity as $x$ tends to 0. Analytically i'm not sure how to continue this. But putting it into Wolfram Alpha yields:

http://www.wolframalpha.com/input/?i=f%28x%2B1%29+-+f%28x%29+%3D+2%5E%7B1-x%7Dlog_2%28x%29

That the expression is dependent on polylogarithms (unsurprising) and something called the lerch transcendent (I believe these have to do with generalizing the riemann zeta function).

So if you choose to go down the finite difference path, it seems that you might be using some heavy analytic number theory tools to determine if the expression is transcendental.

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protected by Community Jun 13 '15 at 21:29

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