Question Number 168776 by safojontoshtemirov last updated on 17/Apr/22
Commented by Oktamboy last updated on 17/Apr/22
$${Stirling} \\ $$
Commented by infinityaction last updated on 17/Apr/22
$${got}\:{it}\: \\ $$$${thank}\:{you}\:{sir} \\ $$
Commented by infinityaction last updated on 17/Apr/22
$${sir}\:{share}\:{your}\:{slution}\:{please} \\ $$
Commented by Oktamboy last updated on 17/Apr/22
$${Stirling}\:{formula}: \\ $$$${n}!\backsim\left(\frac{{n}}{{e}}\right)^{{n}} ×\sqrt{\mathrm{2}{pi}×{n}} \\ $$
Answered by qaz last updated on 18/Apr/22
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{n}}]{\frac{\mathrm{3}^{\mathrm{3n}} \left(\mathrm{n}!\right)^{\mathrm{3}} }{\left(\mathrm{3n}\right)!}}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{\mathrm{3n}} \left(\mathrm{n}!\right)^{\mathrm{3}} }{\left(\mathrm{3n}\right)!}\centerdot\frac{\left(\mathrm{3}\left(\mathrm{n}−\mathrm{1}\right)\right)!}{\mathrm{3}^{\mathrm{3}\left(\mathrm{n}−\mathrm{1}\right)} \left(\left(\mathrm{n}−\mathrm{1}\right)!\right)^{\mathrm{3}} } \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} }{\mathrm{3n}\left(\mathrm{3n}−\mathrm{1}\right)\left(\mathrm{3n}−\mathrm{2}\right)}=\mathrm{1} \\ $$