Question

${lim}_{n\to \infty }(\frac{1}{n}+\frac{e^{\frac{1}{n}}}{n}+\frac{e^{\frac{2}{n}}}{n}+...+\frac{e^{\frac{n-1}{n}}}{n})$ equals

• A. $0$
• B. $1$
• C. $e-1$
• D. $e$

Answer 1

The correct option is C $e-1$

$={lim}_{n\to \infty }\frac{1}{n}+\frac{e^{\frac{1}{n}}}{n}+\frac{e^{\frac{2}{n}}}{n}+\frac{e^{\frac{3}{n}}}{n}+...+\frac{e^{\frac{n-1}{n}}}{n}$

$={lim}_{n\to \infty }\frac{1}{n}[1+e^{\frac{1}{n}}+e^{\frac{2}{n}}+e^{\frac{3}{n}}+...+e^{\frac{n-1}{n}}]$

$={lim}_{n\to \infty }\frac{1}{n}+{lim}_{n\to \infty }\frac{1}{n}[e^{\frac{1}{n}}+e^{\frac{2}{n}}+e^{\frac{3}{n}}+...+e^{\frac{n-1}{n}}]$

$=0+{lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^n{e}^{\frac{r}{n}}$ by using the definition of integral we have

$={\int }_0^1{e}^{x}dx={\left[e^x\right]}_0^1=e^1-e^0=e-1$