A) 0
B) \[{{\log }_{e}}4\]
C) \[{{\log }_{e}}3\]
D) \[{{\log }_{e}}2\]
Correct Answer: D
Solution :
\[\underset{n\to \infty }{\mathop{\text{lim}}}\,\left[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+.....+\frac{1}{2n} \right]\] = \[\underset{n\to \infty }{\mathop{\text{lim}}}\,\,\left[ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{n+n} \right]\] \[=\frac{1}{n}\underset{n\to \infty }{\mathop{\text{lim}}}\,\left[ 1+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+....+\frac{1}{1+\frac{n}{n}} \right]\] \[=\frac{1}{n}\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{r=0}^{n}{\left[ \frac{1}{1+\frac{r}{n}} \right]}\] \[=\int_{0}^{1}{\frac{1}{1+x}\,\,dx}\] \[=[{{\log }_{e}}(1+x)]_{0}^{1}={{\log }_{e}}2-{{\log }_{e}}1={{\log }_{e}}2\].
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