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)
$\lim_{n \rightarrow \infty}{\sum_{r = 1}^{n}{\frac{1}{n}e^{r/n} = \int_{0}^{1}{e^{x}dx = \left\lbrack e^{x} \right\rbrack_{0}^{1}}}}$
$= e - 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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