Q. $\lim_{n \rightarrow \infty}{\sum_{r = 1}^{n}{\frac{1}{n}e^{r/n}}}$ is
A) $e$
B) $e - 1$
C) $1 - e$
D) $e + 1$
Explanation
(b)
$\lim_{n \rightarrow \infty}{\sum_{r = 1}^{n}{\frac{1}{n}e^{\frac{r}{n}}}} = \int_{0}^{1}{e^{x}dx} = \left\lbrack e^{x} \right\rbrack_{0}^{1} = e - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.