How to find the value of $\sqrt{1\sqrt{2\sqrt{3 \cdots}}}$? I thought up this question recently, and I think I've figured out the partial sum:
$$
S_n := \left(n\prod_{k=2}^{n-1} k^{2^{n-k}}\right)^{2^{-k}}.
$$
But I don't even quite know if I'm on the right track. If I am, how do I find the limit of the above equation, and if not, how can I find it another way? Thanks.
 A: This is Somos' quadratic recurrence constant $($see also$)$, whose value is about $1.661688^{^-}$ and which is not yet known to possess a closed form. For a similar expression, see the nested radical constant. I think this should help out.
A: We see that
$$
S_n = 1^{\frac{1}{2}}2^{\frac{1}{4}}3^{\frac{1}{8}}\ldots = \prod_{k=1}^{n}k^{2^{-k}}.
$$
Furthermore
$$
\log S_n = \sum_{k=1}^{n}\log k^{2^{-k}} = \sum_{k=1}^{n}\left(\frac{1}{2}\right)^k\log k.
$$
So we see that 
$$
\lim_{n\rightarrow\infty}\log S_n = \sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\log k \le \sum_{k=1}^{\infty}k\left(\frac{1}{2}\right)^k = \\\frac{1}{2}\sum_{k=1}^{\infty}k\left(\frac{1}{2}\right)^{k-1} = \frac{1}{2}\frac{1}{(1-1/2)^2} = \frac{1}{8}<\infty.
$$
So the product definitely converges. Unfortunately I don't know about the exact limit.
A: An easy way of showing that the product is bounded, is to replace the sequence with $2^n$ instead of $n$.
The product is less that $ 2^{\frac{1}{2}} \times 2^{\frac{2}{4} } \times 2^{ \frac{3}{8} } \times 2^{ \frac{ 4} { 16} } \ldots  = 2^2 = 4$.
