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I'm new here and I'm wondering what $ \frac{1}{2}^{\frac{1}{3}^\frac{1}{4}}$... to infinity is. I don't know either the recursive formula or the generating formula, nor the limit itself. Anyone have any ideas?

P.S. I mean tetration, like (1/2)^((1/3)^(1/4))..., not 1/2^(the product of all 1/n from 3 to infinity)

P.S.S. I came up with this sequence just for fun and got stuck on it. I decided to calculate the approximate sequence in Desmos, which was about 0.65-0.69. And ChatGPT doesn’t understand me at all, so I’m asking you how I can find the exact value of this number to infinity. Because my sequence doesn’t go to 1 at all.

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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Dec 31 '24 at 17:24
  • I would call it iterative, rather than recursive. – Weather Vane Dec 31 '24 at 17:29
  • If $0 < a < 1$ then $0< a < a^{\frac 12} < a^{\frac 13} < a^{\frac 14} < .... < 1$. (Do you know that?). And $\lim_{k\to\infty} a^{\frac 1k} =1$. (Do you know that). So.... $a^{\frac 1{k+1}} < ({{{a^{\frac 12}}^{\frac 13}}^{....}}^{\frac 1k})^{\frac 1{k+1}}$ while $a^{\frac 1k}\to 1$. Can you take it from there? – fleablood Dec 31 '24 at 17:33
  • A similar number : https://oeis.org/A102575 – mick Jun 04 '25 at 21:42

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