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Closed form for $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$

This is equivalent to $\prod_{n=1}^\infty n^{1/2^n}$.

Putting this into Wolframalpha gives that it is approximately 1.661687, and failed to find a closed form.

(1) Is this irrational and transcendental, irrational and algebraic, or rational?

(2) Is there a name for this constant or does there exist a possible closed form?

(3) How does one calculate its partial sum formula? Wolfram Partialsum

Teoc
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  • Putting that number into a search engine gives this or this. – mvw Dec 09 '14 at 02:57
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    The second link is merely the same question as this one, just with a log taken to make the product a sum. The first link does not match with the value given. – Teoc Dec 09 '14 at 03:03

2 Answers2

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Yes, there is a name for it. It is called the Somos' Quadratic Recurrence Constant. It has a weird closed-form in terms of the polylogarithm and Lerch transcendent.

P.S. You can search for constants in the OEIS using its decimal expansion. If you're lucky, then it might be there like in this case: 1,6,6,1,6,8, (A112302)

Tito Piezas III
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    Sigh, did you look at the definition of https://oeis.org/A112302? It is precisely $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$ – Tito Piezas III Dec 09 '14 at 03:06
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    Wolfram may be wrong. Check out the Mathworld Link. – Cheerful Parsnip Dec 09 '14 at 03:07
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    Wolfram alpha and mathematica are not infallible. – Cheerful Parsnip Dec 09 '14 at 03:08
  • Is it irrational and transcendental? Mathworld does not have a notice on that. – Teoc Dec 09 '14 at 03:11
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    @MathNoob: We are pretty sure it is transcendental, but probably don't have a proof. Unless a number has a reason to be algebraic, it is transcendental-there are only countably many algebraic numbers. This doesn't seem to have a reason to be algebraic. – Ross Millikan Dec 09 '14 at 04:57
  • @RossMillikan Although this argument is often used (also for irrationality or normality) , we should not claim a number to be XY if we cannot prove it (just because we cannot determine a minimal polynomial). Upto some degree , we can find a lower bound for the absolute value of the coefficients of a possible minimal polynomial , if this is high upto a high degree, we have "a good chance" to have a transcendental number. – Peter Oct 31 '22 at 08:44
  • Moreover, the constant problem might prevent us to establish the given number to be algebraic , should this actually be the case. – Peter Oct 31 '22 at 08:46
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Let $$A=\prod_{n=1}^\infty n^{1/2^n}$$ $$\log(A)=\sum_{n=1}^\infty\frac{1}{2^n}\log(n)$$ and, from a CAS, the result is $$\log(A)=-\text{PolyLog}^{(1,0)}\left(0,\frac{1}{2}\right)$$ $$A=e^{-\text{PolyLog}^{(1,0)}\left(0,\frac{1}{2}\right)}$$ where appears a derivative of the polylogarithm function.

Claude Leibovici
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