Question Number 121716 by bemath last updated on 11/Nov/20
$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)…\left(\mathrm{3}{n}\right)}{{n}^{\mathrm{2}{n}} }\:\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$
Answered by Dwaipayan Shikari last updated on 11/Nov/20
![lim_(n→∞) ((1+(1/n))(1+(2/n))...(1+((2n)/n)))^(1/n) =y lim_(n→∞) (1/n)Σ_(r=1) ^(2n) log(1+(r/n))=logy ∫_0 ^2 log(1+x)dx=logy [xlog(1+x)−x+log(x+1)]_0 ^2 =logy 3log3−2=logy y=((27)/e^2 )](https://www.tinkutara.com/question/Q121718.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{\mathrm{2}{n}}{{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} ={y} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}{log}\left(\mathrm{1}+\frac{{r}}{{n}}\right)={logy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} {log}\left(\mathrm{1}+{x}\right){dx}={logy} \\ $$$$\left[{xlog}\left(\mathrm{1}+{x}\right)−{x}+{log}\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} ={logy} \\ $$$$\mathrm{3}{log}\mathrm{3}−\mathrm{2}={logy} \\ $$$${y}=\frac{\mathrm{27}}{{e}^{\mathrm{2}} } \\ $$