Is is possible to use Arithmetic progression in this case?

**xcluded** · October 22nd 2009, 07:36 PM

$n$
$\sum e^\frac{i}{n}(\frac{1}{n})$
$i = 1$

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

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

Thanks.

**ramiee2010** · October 22nd 2009, 10:26 PM

Originally Posted by xcluded

$n$
$\sum e^\frac{i}{n}(\frac{1}{n})$
$i = 1$

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

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

Thanks.

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

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