# Sum of this geometric series

• Feb 2nd 2013, 01:36 PM
director
Sum of this geometric series
Hi all,

I'm trying to find:
$\sum_{k=0}^{n} (e^{i\Theta})^{k}$

I manipulate this the following way:

$= 1 + \sum_{k=1}^{n} (e^{i\Theta})^{k} = 1 + \frac{1-e^{in\Theta}}{1-e^{i\Theta}}$

Is this right? Thanks
• Feb 2nd 2013, 02:58 PM
Shakarri
Re: Sum of this geometric series
Yes you can derive that result directly in the same way that the general formula is derived, I just tried it myself and the peculiar common ratio in the series does not change the derivation.
• Feb 2nd 2013, 03:10 PM
director
Re: Sum of this geometric series
I see, thanks. Now, If I want to obtain the sum from just the real part of $(e^{i\Theta})^{k}$,
that is, $\sum_{k=0}^{n} cos(k\Theta)$

Could I start from:

$\sum_{k=0}^{n} (e^{i\Theta})^{k} = \sum_{k=0}^{n} cos(k\Theta) + i\sum_{k=0}^{n} sin(k\Theta)$

and then continue this way:

$\sum_{k=0}^{n} cos(k\Theta) = \sum_{k=0}^{n} (e^{i\Theta})^{k} - i\sum_{k=0}^{n} sin(k\Theta)$

Or is there an easier method?
• Feb 5th 2013, 03:49 AM
ibdutt
Re: Sum of this geometric series
The expression for sum of n terms of a GP is
Attachment 26861 Now I am sure you can make out where did you go wrong
• Feb 5th 2013, 03:53 AM
Prove It
Re: Sum of this geometric series
Actually it's \displaystyle \begin{align*} S_n = \frac{a\left( 1 - r^n \right)}{1 - r} \end{align*}...