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

Question Number 142464 by som(math1967) last updated on 01/Jun/21

$$\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\:\left[\frac{\left(\mathrm{2}\boldsymbol{{n}}\right)!}{\boldsymbol{{n}}!\boldsymbol{{n}}^{\boldsymbol{{n}}} }\right]^{\frac{\mathrm{1}}{\boldsymbol{{n}}}} =? \\ $$

Answered by Dwaipayan Shikari last updated on 01/Jun/21

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{2}^{\mathrm{2}{n}} {n}^{\mathrm{2}{n}} {e}^{−\mathrm{2}{n}} \sqrt{\mathrm{4}\pi{n}}}{{n}^{{n}} {e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:{n}^{{n}} }\right]^{\frac{\mathrm{1}}{{n}}} ={y} \\ $$$$\Rightarrow\left[\left(\frac{\mathrm{4}}{{e}}\right)^{{n}} \sqrt{\mathrm{2}}\right]^{\frac{\mathrm{1}}{{n}}} ={y}\Rightarrow{y}=\frac{\mathrm{4}}{{e}} \\ $$

Answered by Dwaipayan Shikari last updated on 01/Jun/21

$$\frac{\mathrm{1}}{{n}}{log}\left(\left(\mathrm{2}{n}\right)!\right)−\frac{\mathrm{1}}{{n}}{log}\left(\frac{{n}!}{{n}^{{n}} }\right)−\frac{\mathrm{1}}{{n}}{log}\left({n}^{\mathrm{2}{n}} \right)={log}\left({y}\right) \\ $$$$\frac{\mathrm{1}}{{n}}{log}\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}^{\mathrm{2}{n}} }\right)−\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{log}\left(\frac{{r}}{{n}}\right)={log}\left({y}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}{log}\left(\frac{{r}}{{n}}\right)−\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{log}\left(\frac{{r}}{{n}}\right)={log}\left({y}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2}} {log}\left({x}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right){dx}={log}\left({y}\right) \\ $$$$\Rightarrow{log}\left(\mathrm{4}\right)−\mathrm{2}+\mathrm{1}={log}\left({y}\right)\Rightarrow{y}=\frac{\mathrm{4}}{{e}} \\ $$

Commented by som(math1967) last updated on 01/Jun/21

$$\boldsymbol{{Thank}}\:\boldsymbol{{you}} \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com