$ lim_{n \rightarrow \infty} (\frac{1}{n} + \frac{e^{1/n}}{n} + \frac{e^{2/n}}{n} + ….+ \frac{e^{n-1/n}}{n}) $ equals

Q: $\large lim_{n \rightarrow \infty} (\frac{1}{n} + \frac{e^{1/n}}{n} + \frac{e^{2/n}}{n} + ….+ \frac{e^{n-1/n}}{n}) $ equals

(A) 0

(B) e

(C) e – 1

(D) e + 1

Required limit = $\large lim_{n \rightarrow \infty} \frac{1}{n}. \frac{(e^{1/n})^n – 1}{e^{1/n} – 1} $

$\large (e-1)lim_{n \rightarrow \infty} [ \frac{1}{\frac{e^{1/n} – 1}{1/n}}] $

= (e – 1) × 1 = e – 1.

Hence (C) is the correct answer.