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

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

• Putting that number into a search engine gives this or this. – mvw Dec 9 '14 at 2:57
• 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 9 '14 at 3:03

• Sigh, did you look at the definition of oeis.org/A112302? It is precisely $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$ – Tito Piezas III Dec 9 '14 at 3:06
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.