Find-lim-n-1-n-2n-n-

Question Number 205885 by hardmath last updated on 01/Apr/24

Find : \lim_{n \to \infty} \sqrt[n]{} (\begin{matrix} 2 n \\ n \end{matrix}) = ?

Answered by MM42 last updated on 03/Apr/24

(\begin{matrix} 2 n \\ n \end{matrix}) = \frac{(2 n) (2 n - 1) (2 n - 2) \dots (n + 3) (n + 2) (n + 1)}{n (n - 1) (n - 2) \dots 3 \times 2 \times 1}

a = \sqrt[n]{} (\begin{matrix} 2 n \\ n \end{matrix})

\Rightarrow l n a = \frac{1}{n} [l n (2) + l n (2 + \frac{1}{n - 1}) + l n (2 + \frac{2}{n - 2}) + \dots + l n (2 + \frac{n - 3}{3}) (2 + \frac{n - 2}{2}) (2 + \frac{n - 1}{1})

= \frac{1}{n} \underset{i = 0}{\sum^{n - 1}} l n (2 + \frac{\frac{i}{n}}{1 - \frac{i}{n}})

\Rightarrow {l i m}_{n \to \infty} a_{n} = \int_{0}^{1} l n (2 + \frac{x}{1 - x}) f x

= \int_{0}^{1} [l n (2 - x) - l n (1 - x)] d x

{= (x - 2) l n (2 - x) - (x - 1) l n (1 - x))]}_{0}^{1}

= 2 l n 2 = l n 4 \Rightarrow {l i m}_{n \to \infty} a_{n} = 4 ✓

Answered by Frix last updated on 01/Apr/24

n \to \infty \Rightarrow n! \to {(\frac{n}{e})}^{n} \sqrt{2 π n} \Rightarrow

\sqrt[n]{(\begin{matrix} 2 n \\ n \end{matrix})} \to \frac{4}{\sqrt[n]{π n}} = 4

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