Sum of this geometric series

**director** · February 2nd 2013, 12:36 PM

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

**Shakarri** · February 2nd 2013, 01:58 PM

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.

**director** · February 2nd 2013, 02:10 PM

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?

**ibdutt** · February 5th 2013, 02:49 AM

The expression for sum of n terms of a GP is

Now I am sure you can make out where did you go wrong

**Prove It** · February 5th 2013, 02:53 AM

Actually it's $\displaystyle \begin{align*} S_n = \frac{a\left( 1 - r^n \right)}{1 - r} \end{align*}$ ...

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