Question
$$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right).....3n}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}}$$ is equal to:
A.
$${\frac{{9}}{{{e^2}}}}$$
B.
$$3\,\log \,3 - 2$$
C.
$${\frac{{18}}{{{e^4}}}}$$
D.
$${\frac{{27}}{{{e^2}}}}$$
Answer :
$${\frac{{27}}{{{e^2}}}}$$
Solution :
$$\eqalign{
& y = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right).....3n}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} \cr
& \ln \,y = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\ln \left( {1 + \frac{1}{n}} \right)\left( {1 + \frac{2}{n}} \right)...\left( {1 + \frac{{2n}}{n}} \right) \cr
& \ln \,y = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\ln \left( {1 + \frac{1}{n}} \right) + \ln \left( {1 + \frac{2}{n}} \right) + ..... + \ln \left( {1 + \frac{{2n}}{n}} \right)} \right] \cr
& = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\ln \left( {1 + \frac{r}{n}} \right)} \cr
& = \int_0^2 {\ln } \left( {1 + x} \right)dx \cr
& {\text{Let }}1 + x = t\,\,\, \Rightarrow dx = dt \cr
& {\text{when }}x = 0,\,\,t = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,\,t = 3 \cr
& \ln \,y = \int_1^3 {\ln } \,t\,d\,t = \left[ {t\,\ln \,t - t} \right]_1^3 = \ln \left( {\frac{{{3^3}}}{{{e^2}}}} \right) = \ln \left( {\frac{{27}}{{{e^2}}}} \right) \cr
& \Rightarrow y = \left( {\frac{{27}}{{{e^2}}}} \right) \cr} $$