Irrationality of √2√3√4⋯\sqrt{2\sqrt{3\sqrt{4\cdots}}}

In this question it is stated that Somos’ quadratic recurrence constant
α=234
is an irrational number. [update: the author of that question is no longer claiming to have a proof of this]

This fact seems by no means trivial to me. The algebraic numbers 2, 23, 234, do not converge quickly enough to α, so one cannot reuse the proof of Liouville’s theorem in this case.

Approximation arguments do not seem a good way, since
222=2
is rational instead!

What am I missing?

Answer

Your original post implies you have thought of, or know of, a good way of using Liouville’s theorem to solve the problem. Here’s the start of some ideas on the issue that perhaps you will be able to finish off.

We can write α in the following way:
α=n=1(1+1n)2(n1).

We will also define the partial product

αN=Nn=1(1+1n)2(n1).

Then, for some integers p, q:

|αpq||ααN|+|αNpq|.

Now, the first of these values can be quickly bounded:

ααN=αN(n=N+1(1+1n)2(n1)1)αN(n=N+1(2)2(n1)1)αN(22(N1)1)4αN2N4α2N

However, I can’t see how to bound the other expression, |αNpq|. Because the αN are algebraic, we can’t instantly point to the existence of some rationals that approximate them “well enough” for our purposes here, as far as I can see.

Attribution
Source : Link , Question Author : Mizar , Answer Author : Mizar

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