Question

The value of $\underset{n\to \infty }{\lim }\left(\frac{1}{n}+\frac{e^{1/n}}{n}+\frac{e^{2/n}}{n}+\dots +\frac{e^{(n-1)/n}}{n}\right)$ is

• A. 1
• B. 0
• C. e-1
• D. e+1

Answer 1

Answer: (C)

$\underset{n\to \infty }{\lim }\left[\frac{1}{n}+\frac{e^{1/n}}{n}+\frac{e^{2/n}}{n}+\dots +\frac{e^{(n-1)/n}}{n}\right]$
$=\underset{n\to \infty }{\lim }\left[\frac{1+e^{1/n}+{\left(e^{1/n}\right)}^2+\dots +{\left(e^{1/n}\right)}^{n-1}}{n}\right]$
$=\underset{n\to \infty }{\lim }\frac{1\cdot \left[{\left(e^{1/n}\right)}^n-1\right]}{n\left(e^{1/n}-1\right)}$
$=(e-1)\underset{n\to \infty }{\lim }\frac{1}{\left(\frac{e^{1/n}-1}{1/n}\right)}$
$=(e-1)\times 1=(e-1)$