A Question Of Convergence

Calculus Level 3

Does the following converge?

lim n e 2 ! e 2 0.5 ln 2 + e 3 ! e 3 0.5 ln 3 + e 4 ! e 4 0.5 ln 4 + n n n \lim_{n \rightarrow \infty} \sqrt[n]{e^{2!e^{2-0.5\ln 2}}+\sqrt[n]{e^{3!e^{3-0.5\ln 3}}+\sqrt[n]{e^{4!e^{4-0.5\ln 4}}+\ldots}}}

All the tests are inconclusive >~< It diverges! It converges! Huh? What's convergence?

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1 solution

Jake Lai
Dec 29, 2014

By Stirling's approximation , we can see that

x n p n = ( e n ! e n 0.5 ln n ) 1 n n = e n ! ( n / e ) n n e 2 π x_{n}^{p^{n}} = (e^{n!e^{n-0.5\ln n}})^{\frac{1}{n^{n}}} = e^{\frac{n!}{(n/e)^{n}\sqrt{n}}} \rightarrow e^{\sqrt{2\pi}}

is bounded; hence, the infinitely nested radicals converge due to Herschfeld's Convergence Theorem .

You are having heavy amount of knowledge truly

U Z - 6 years, 5 months ago

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I honestly just guessed it (correct).

Steven Zheng - 6 years, 5 months ago

It's easy to guess, but not to prove. This question was actually inspired by Michael Mendrin in his response to whether or not 1 + 2 + 3 + \sqrt{1+\sqrt{2+\sqrt{3+\ldots}}} converges.

Jake Lai - 6 years, 5 months ago

But I think for any finite n n the nested radical itself is inf! For any finite n n think how you define that expression. The value of that expression should at least as big as e m ! / n m e^{m!/n^m} for all m N m\in\mathbb{N} which clearly diverge to infinity as m m goes to infinity

Cuize Han - 6 years, 5 months ago

The link to Herschfeld's Theorem states the result for a fixed power (root) p. The exponents in the problem are decreasing with the index.

Mathew Titus - 6 years, 4 months ago

I feel the .... in this problem is a tad ambiguous. I presume there are n terms, but that could be clearer.

Joe Mansley - 2 years, 6 months ago

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