Question

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

• A. $1$
• B. $0$
• C. $e-1$
• D. $e+1$

Answer 1

The correct option is C $e-1$

$\underset{n\to \infty }{lim}\left[\frac{1}{n}+\frac{e^{1/n}}{n}+\frac{e^{2/n}}{n}+...+\frac{e^{(n-1)/n}}{n}\right]$
$=\underset{n\to \infty }{lim}\left[\frac{1+e^{1/n}+(e^{1/n}{)}^2+...+(e^{1/n{)}^{n-1}}}{n}\right]$
$=\underset{n\to \infty }{lim}\frac{1.\left[\right(e^{1/n}{)}^n-1]}{n(e^{1/n}-1)}$$=(e-1)\underset{n\to \infty }{lim}\frac{1}{\left(\frac{e^{1/n}-1}{1/n}\right)}$
$=(e-1)\times 1=(e-1)$
Hence, option 'C' is correct.