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Question: \[\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + .....\fr...

lim⁡n→∞[1n+1n+1+1n+2+.....12n]=\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + .....\frac{1}{2n} \right\rbrack =

A

0

B

log⁡e4\log_{e}4

C

log⁡e3\log_{e}3

D

log⁡e2\log_{e}2

Answer

log⁡e2\log_{e}2

Explanation

Solution

limn→∞[1n+1n+1+1n+2+.....+12n]\underset{n \rightarrow \infty}{\text{lim}}\left\lbrack \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + ..... + \frac{1}{2n} \right\rbrack

= limn→∞[1n+1n+1+1n+2+....+1n+n]\underset{n \rightarrow \infty}{\text{lim}}{}\left\lbrack \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + .... + \frac{1}{n + n} \right\rbrack

=1nlimn→∞[1+11+1n+11+2n+....+11+nn]= \frac{1}{n}\underset{n \rightarrow \infty}{\text{lim}}\left\lbrack 1 + \frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + .... + \frac{1}{1 + \frac{n}{n}} \right\rbrack

=1nlimn→∞∑r=0n[11+rn]= \frac{1}{n}\underset{n \rightarrow \infty}{\text{lim}}{\sum_{r = 0}^{n}\left\lbrack \frac{1}{1 + \frac{r}{n}} \right\rbrack} =∫0111+xdx= \int_{0}^{1}{\frac{1}{1 + x}dx}

=[log⁡e(1+x)]01=log⁡e2−log⁡e1=log⁡e2= \lbrack\log_{e}(1 + x)\rbrack_{0}^{1} = \log_{e}2 - \log_{e}1 = \log_{e}2.