Neperando

[apritisesamo]

$\fs{4}\text \blue {E' vero che } \lim_{n\to \infty} \frac{1}{n}\sqr[n]{\frac{(2n)!}{n!}} = \frac{4}{e} \fs{6}\text { ?}$

[panurgo]

Sì!

[panurgo]

Sì!

$\begin{array}{lC+50} \lim_{n\/\to\/\infty} \frac{1}{n}\sqr[n]{\frac{\left(2n\right)!}{n!}}\/=\/\frac{4}{e} \\ \log\left\[ \lim_{n\/\to\/\infty} \frac{1}{n}\sqr[n]{\frac{\left(2n\right)!}{n!}}\right\]\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\log\left\{ \frac{1}{n}\left\[{\frac{\left(2n\right)!}{n!}}\right\]^{\script \frac1n}\right\}\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\left\{\frac1n\/ \log\left\[{\frac{\left(2n\right)!}{n!}}\right\]-\log n\right\}\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\left\{\frac1n\/ \left\[{\log\left(2n\right)!\/-\/\log{n!}}\right\]-\log n\right\}\/=\/2\/\log2\/-\/1 \\ \log{n!}\/\sim\/\frac12\/\log\left(2\pi n\right)\/+\/n\/\log{n}\/-\/n \\ \lim_{n\/\to\/\infty}\left\{\frac1n\/ \left\[\frac12\/\log\left(4\pi n\right)\/+\/2n\/\log{2n}\/-\/2n\/-\/\frac12\/\log\left(2\pi n\right)\/-\/n\/\log{n}\/+\/n\right\]-\log n\right\}\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\left\{\frac1n\/ \left\[\frac12\/\log\left(4\pi n\right)\/-\/\frac12\/\log\left(2\pi n\right)\/+\/2n\/\log{2n}\/-\/2n\/\log{n}\/-\/2n\/+\/n\right\]\right\}\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\left\{\frac1n\/ \left\[\frac12\/\log\frac{4\pi n}{2\pi n}\/+\/2n\/\log\frac{2n}{n}\/-\/n\right\]\right\}\/=\/2\/\log2\/-\/1 \\ \lim_{n\/\to\/\infty}\left\[\frac1{2n}\/\log2\right\]\/+\/2\/\log2\/-\/1\/=\/2\/\log2\/-\/1 \\ {\text Q.E.D.} \end{array}$

[karl]

Ottima soluzione...Io ero riuscito solo a dimostrare che il limite L richiesto soddisfa la relazione:
$\large \sqrt{2}<L<\frac{3}{2}$