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I am given a sequence whose limit, as n approaches infinity is 1. I want to calculate the limit of another sequence based off that.

The given sequence is: $\lim_{n\to\infty} (\frac{n!}{n^n e^{-n} \sqrt{2\pi n}}) = 1$

The limit I want to find is for the following sequence: $\lim_{n\to\infty} (\frac{(2n)!\sqrt{n}}{(n!)^2 4^n}) $

Thanks

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Put $$ a_n := \frac{n!}{n^ne^{-n}\sqrt{2\pi n}} \quad\text{and}\quad b_n := \frac{(2n)!\sqrt{n}}{4^n(n!)^2} $$ Then \begin{align*} a_{2n} = \frac{(2n)!}{2(2n)^{2n}e^{-2n}\sqrt{\pi n}} = \ldots = \frac{(2n)!\sqrt{n}}{4^n(n!)^2}\cdot\frac{(n!)^2}{2n^{2n+1}e^{-2n}\sqrt{\pi}} = b_n\cdot a_n^2\sqrt\pi. \end{align*} Letting $n\to\infty$ gives $\lim_{n\to\infty}b_n = \frac 1{\sqrt\pi}$.

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