$\sqrt{2\sqrt{3\sqrt{4.....\infty}}}$

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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!

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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.