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How can i find the value of $\sqrt{2\sqrt{3\sqrt{4.....\infty}}}$? I had problem in this question because this is not like the question $\sqrt{2\sqrt{2\sqrt{2...\infty}}}$ where the numbers are symmetrical on the other hand, $\sqrt{2{\sqrt{3\sqrt{4.....\infty}}}}$ has numbers in increasing order. I was not able to make any progress in this question, i have no idea to how i can solve this question. Please help me with this problem!

Mathzcreator
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    Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. – saulspatz Apr 24 '21 at 04:00
  • See this question. It is quite similar to yours. – soupless Apr 24 '21 at 04:04
  • @Mathzcreator Based from an answer there, you can get what you're looking for. Just a few adjustments will do. – soupless Apr 24 '21 at 04:12
  • See A259235. – g.kov Apr 24 '21 at 04:15
  • @Mathzcreator Your expression is just the square of Somos's Quadratic Recurrence Constant, from the answer linked to by soupless. Wolfram Math World knows of no nice expression in terms of elementary functions, so unless someone comes up with amazing original research on the spot, you're not going to get an answer nicer than that. – Theo Bendit Apr 24 '21 at 05:59

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See Sosmos's Quadratic Recurrence Constant. Your nested radical is equal to $$\sqrt{2\sqrt{3\sqrt{4\sqrt{...}}}}=\sigma^2\approx 2.76120684185$$ where $\sigma$ is Sosmos's Quadratic Recurrence Constant.