Question

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

• A. $e$
• B. $-e$
• C. $e-1$
• D. $1-e$

Answer 1

Answer: (C)

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