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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

$$gotit$$$$thankyousir$$

Commented by infinityaction last updated on 17/Apr/22

$$sirshareyourslutionplease$$

Commented by Oktamboy last updated on 17/Apr/22

$$Stirlingformula:$$$$n!\backsim {\left(\frac{n}{e}\right)}^n\times \sqrt{2pi\times n}$$

Answered by qaz last updated on 18/Apr/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathrm{n}]{\frac{3^{3\mathrm{n}}{(\mathrm{n}!)}^3}{\left(3\mathrm{n}\right)!}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{3^{3\mathrm{n}}{(\mathrm{n}!)}^3}{\left(3\mathrm{n}\right)!}\cdot \frac{\left(3(\mathrm{n}-1)\right)!}{3^{3(\mathrm{n}-1)}{((\mathrm{n}-1)!)}^3}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{3^3{\mathrm{n}}^3}{3\mathrm{n}(3\mathrm{n}-1)(3\mathrm{n}-2)}=1$$

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