Hi all,

I'm trying to find:

$\displaystyle \sum_{k=0}^{n} (e^{i\Theta})^{k}$

I manipulate this the following way:

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

Is this right? Thanks

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- Feb 2nd 2013, 12:36 PMdirectorSum of this geometric series
Hi all,

I'm trying to find:

$\displaystyle \sum_{k=0}^{n} (e^{i\Theta})^{k}$

I manipulate this the following way:

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

Is this right? Thanks - Feb 2nd 2013, 01:58 PMShakarriRe: 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, 02:10 PMdirectorRe: Sum of this geometric series
I see, thanks. Now, If I want to obtain the sum from just the real part of $\displaystyle (e^{i\Theta})^{k}$,

that is, $\displaystyle \sum_{k=0}^{n} cos(k\Theta)$

Could I start from:

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

and then continue this way:

$\displaystyle \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, 02:49 AMibduttRe: 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, 02:53 AMProve ItRe: Sum of this geometric series
Actually it's $\displaystyle \displaystyle \begin{align*} S_n = \frac{a\left( 1 - r^n \right)}{1 - r} \end{align*}$...