Definite Integration Question 207
Question: $ \underset{n\to \infty }{\mathop{\lim }}{{
[ \frac{n!}{n^{n}} ]}^{1/n}} $ equals [Kurukshetra CEE 1998]
Options:
A) e
B) $ 1/e $
C) $ \pi /4 $
D) $ 4/\pi $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ P=\underset{x\to \infty }{\mathop{\lim }}{{( \frac{n!}{n^{n}} )}^{1/n}} $
$ =\underset{n\to \infty }{\mathop{\lim }}{{( \frac{1}{n}.\frac{2}{n}.\frac{3}{n}.\frac{4}{n}……….\frac{n}{n} )}^{1/n}} $
$ \therefore \log P=\frac{1}{n}\underset{n\to \infty }{\mathop{\lim }}( \log \frac{1}{n}+\log \frac{2}{n}+……+\log \frac{n}{n} ) $
$ \log P=\underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{{}}\frac{1}{n}\log \frac{r}{n} $
$ \log P=\int_0^{1}{{}}\log xdx=(x\log x-x)_0^{1}=(-1) $
therefore $ P=\frac{1}{e} $ .
BETA
