$\sum_{i=0}^{2n-1} e^\frac{i}{n}$?

Question

I was doing a Riemann integral of $\int_0^2e^xdx$ and ended up with

$\overline S(\mathcal{D})= \frac{e^{\frac{1}{n}}}{n} \sum_{i=0}^{2n-1}e^\frac{i}{n}$.

I was wondering how you would compute this sum. Thanks!

Answer 1

The sum of the geometric progression $a,ar,ar^2,...,ar^{m-1}$ is $a\Big[\frac{r^m-1}{r-1}\Big], r\ne1$

$\displaystyle\sum_{i=0}^{2n-1}e^{\frac in}=e^\frac0n+e^\frac1n+e^\frac2n+...+e^\frac{2n-1}n$ is a geometric progression with $a=e^\frac0n=1,r=e^\frac1n$

$\displaystyle\sum_{i=0}^{2n-1}e^{\frac in}=\frac{(e^\frac1n)^{2n}-1}{e^\frac1n-1}=\frac{e^2-1}{e^{\frac1n}-1}$