# Is is possible to use Arithmetic progression in this case?

• Oct 22nd 2009, 07:36 PM
xcluded
Is is possible to use Arithmetic progression in this case?
$\displaystyle n$
$\displaystyle \sum e^\frac{i}{n}(\frac{1}{n})$
$\displaystyle i = 1$

$\displaystyle \frac{1}{n}[ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$

Did i compute wrongly ? I find it a bit weird. I can't proceed further on combining the bracket terms.

Thanks.
• Oct 22nd 2009, 10:26 PM
ramiee2010
Quote:

Originally Posted by xcluded
$\displaystyle n$
$\displaystyle \sum e^\frac{i}{n}(\frac{1}{n})$
$\displaystyle i = 1$

$\displaystyle \frac{1}{n}[ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$

Did i compute wrongly ? I find it a bit weird. I can't proceed further on combining the bracket terms.

Thanks.

$\displaystyle [ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$
it is in geometric progression with common ratio
$\displaystyle = e^\frac{1}{n}$
use sum of infinite G.P. terms formula
$\displaystyle = \frac{a}{1-r}$