# Question be944

Dec 20, 2016

If we expand the summation as follows

$\frac{1}{n} {\sum}_{k = 0}^{n - 1} {e}^{\frac{k}{n}} = \frac{1}{n} \left\{{e}^{\frac{0}{n}} + {e}^{\frac{1}{n}} + {e}^{\frac{2}{n}} + {e}^{\frac{3}{n}} + \ldots + {e}^{\frac{n - 1}{n}}\right\}$
" " = 1/n { underbrace(e^0+e^(1/n)+(e^(1/n))^2 + (e^(1/n))^3 + ... + (e^(1/n))^(n-1))_("n terms") } #

So you are correct, It is a GP with;

first term $a = {e}^{0} \left(= 1\right)$ and
common ratio $r = {e}^{\frac{1}{n}}$

So Using the GP summation formula:

${S}_{n} = a \frac{\left(1 - {r}^{n}\right)}{\left(1 - r\right)}$

to get:

$\frac{1}{n} {\sum}_{k = 0}^{n - 1} {e}^{\frac{k}{n}} = \frac{1}{n} 1 \frac{\left(1 - {\left({e}^{\frac{1}{n}}\right)}^{n}\right)}{\left(1 - {e}^{\frac{1}{n}}\right)}$
$\text{ } = \frac{\left(1 - {e}^{\frac{n}{n}}\right)}{n \left(1 - {e}^{\frac{1}{n}}\right)}$
$\text{ } = \frac{\left({e}^{\frac{n}{n}} - 1\right)}{n \left({e}^{\frac{1}{n}} - 1\right)}$
$\text{ } = \frac{\left({e}^{\frac{n}{n}} - {e}^{0}\right)}{n \left({e}^{\frac{1}{n}} - 1\right)}$ (as ${e}^{0} = 1$) QED