$lim n \to \infty {[\frac{(2 n)!}{n! n^{n}}]}^{1 / n} = ?$

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Solution

We have,
$\begin{matrix} {lim}_{x \to \infty} {[\frac{(2 n)!}{n! n^{n}}]}^{1 / n} = {lim}_{x \to \infty} {[\frac{(2 n) (2 n - 1) . . . . . (n + 1) (n!)}{n! n^{n}}]}^{1 / n} = {lim}_{x \to \infty} {⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{n^{n} (2) (2 - \frac{1}{n}) (2 - \frac{2}{n}) . . . . . (1 + \frac{1}{n}) (n!)}{n! n^{n}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦}^{1 / n} = {lim}_{x \to \infty} {[2^{n}]}^{\frac{1}{n}} = {[2]}^{n \times \frac{1}{n}} = 2 \end{matrix}$

Hence, this is the answer.

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