Question

${lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}]=$ [Karnataka CET 1999]

• A. $0$
• B. $lo{g}_{e}4$
• C. $lo{g}_{e}3$
• D. $lo{g}_{e}2$

Answer 1

The correct option is D $lo{g}_{e}2$

${lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}]$
$={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{n+n}]$
$=\frac{1}{n}{lim}_{n\to \infty }[1+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots +\frac{1}{1+\frac{n}{n}}]$
$=\frac{1}{n}{lim}_{n\to \infty }{\sum }_{r=0}^n\left[\frac{1}{1+\frac{r}{n}}\right]={\int }_0^1\frac{1}{1+x}dx$
$=\left[lo{g}_e\right(1+x){]}_0^1=lo{g}_{e}2-lo{g}_{e}1=lo{g}_{e}2$