Question Number 22871 by keshavnimgade85@gmail.com last updated on 23/Oct/17
![If f(n)=(1/n)[(n+1)(n+2)(n+3)...(n+n)]^(1/n) , then lim_(n→∞) f(n) =](Q22871.png)
$$\mathrm{If}\:\:{f}\left({n}\right)=\frac{\mathrm{1}}{{n}}\left[\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)...\left({n}+{n}\right)\right]^{\mathrm{1}/{n}} , \\ $$$$\mathrm{then}\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right)\:= \\ $$
Answered by ajfour last updated on 23/Oct/17
![f(n)=[(1+(1/n))(1+(2/n))...(1+(n/n))]^(1/n) lim_(n→∞) (ln [f(n)])=∫_0 ^( 1) ln (1+x)dx =xln (1+x)∣_0 ^1 −∫_0 ^( 1) (x/(1+x)) dx =ln 2−[x−ln (1+x)]∣_0 ^1 =ln 2−1+ln 2 =2ln 2−1 ⇒ lim_(n→∞) f(n) =e^(ln (4/e)) =(4/e) .](Q22875.png)
$${f}\left({n}\right)=\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)...\left(\mathrm{1}+\frac{{n}}{{n}}\right)\right]^{\mathrm{1}/{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{ln}\:\left[{f}\left({n}\right)\right]\right)=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \mathrm{ln}\:\left(\mathrm{1}+{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\:\mathrm{2}−\left[{x}−\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\right]\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\:\mathrm{2}−\mathrm{1}+\mathrm{ln}\:\mathrm{2}\:=\mathrm{2ln}\:\mathrm{2}−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right)\:={e}^{\mathrm{ln}\:\left(\mathrm{4}/{e}\right)} \:=\frac{\mathrm{4}}{\boldsymbol{{e}}}\:. \\ $$