Question

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

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

Answer 1

Answer: (C)

$e-1$

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