Definite Integration Question 209
Question: $ \underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+…..\frac{1}{2n} ]= $
[Karnataka CET 1999]
Options:
A) 0
B) $ {\log _{e}}4 $
C) $ {\log _{e}}3 $
D) $ {\log _{e}}2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{n\to \infty }{\mathop{lim}}[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+…..+\frac{1}{2n} ] $
= $ \underset{n\to \infty }{\mathop{lim}}[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+….+\frac{1}{n+n} ] $
$ =\frac{1}{n}\underset{n\to \infty }{\mathop{lim}}[ 1+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+….+\frac{1}{1+\frac{n}{n}} ] $
$ =\frac{1}{n}\underset{n\to \infty }{\mathop{lim}}\sum\limits _{r=0}^{n}{[ \frac{1}{1+\frac{r}{n}} ]} $
$ =\int_0^{1}{\frac{1}{1+x}dx} $
$ =[{\log _{e}}(1+x)]_0^{1}={\log _{e}}2-{\log _{e}}1={\log _{e}}2 $ .
BETA
