${\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}{e}^{\frac{r}{n}}$ is

This limit of a sum can be expressed as a definite integral.
Using the conversion ${\lim }_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}f(\frac{r}{n})={\int }_0^{1}f(x)dx$:
The expression becomes:
${\int }_0^1{e}^{x}dx$.
Evaluating the integral:
$[e^x{]}_0^1=e^1-e^0=e-1$.