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I think this one goes to section of nested radicals, I was trying to solve if for a couple of days now. Maybe you have some nice solution to this one.


I think it should converge. I computed with 1000 elements and 100000000 elements and got same result of 1.66169.

A more general case: $$g(n)=\sqrt{n\sqrt{(n+1)\sqrt{(n+2)\sqrt{(n+3)\sqrt{...}}}}}$$

Has a very nice result of: $$g(n+1)=\frac{g^2(n)}{n}$$

So solving simple case $g(1)$ would also solve more general one.

Edit: My question was if there was a smart way of calculating $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}}$.

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What is your question? – Ali Caglayan Aug 22 '14 at 11:38
My guess is that the OP is asking for the limit $$\lim_{n\to\infty}\left(2^{1/4}3^{1/8}\cdots n^{2^{-n}}\cdots\right)$$ – Quang Hoang Aug 22 '14 at 11:41
some times your intuition of convergence may be wrong. For example: calculate the $1000000000000$ terms of the sequence $s_n=1+1/2+1/3+...+1/n$ – Nilan Aug 22 '14 at 11:43
Or worse: do the same with primes: $s_n=1/2+1/3+1/5+1/7+1/11+\dots+1/p_n$. – Joonas Ilmavirta Aug 22 '14 at 11:45
Duplicate of this. – Lucian Aug 22 '14 at 13:55

2 Answers 2

This is Somos' quadratic recurrence constant $($see also$)$, for which no closed form is yet known. Its decimal expansion can be found here. A similar expression, also lacking a known closed form, is the nested radical constant, whose digits can also be found on OEIS. Hope this helps.

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A similar question can be found here. – Lucian Aug 22 '14 at 16:31

Since $g(1) = 1^{1/2} 2^{1/4} 3^{1/8} \cdots$, we can write $\log g(1)$ as an infinite sum. $$ \log g(1) = \sum_{k=1}^{\infty} \frac{\log k}{2^k}. $$

This series obviously converges since $\log k < 1.5^k$, but I doubt it would be expressed in terms of elementary functions. (See this)

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