# LTI: Is my calculation of the frequency response correct?

i am not sure if my computation is correct. Hope someone can have a look at it. Here is my problem:

\begin{aligned} H[e^{j\omega}] & = \sum_{m=0}^{\infty} (\frac{1}{2} e^{-j\omega})^n = \frac{1}{1-0.5e^{-j\omega}}\\ H[e^{j\omega}] & = \frac{1}{(1.25 - cos{\omega})^{\frac{1}{2}}}\\ \end{aligned}

I got to my solution by looking at a similar example.
But i don't know how they got rid of the sum.

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The summation is simply a geometric series. You remove the summation by using the formula

$$1+x+x^2 +\ldots = \frac{1}{1-x}$$

For the magnitude response, you just have to use your formula $|c|^2 = cc^*$. If $c = a+ib$, this gives you $|c| = \sqrt{a^2 + b^2}$.

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Thank you, now i understand that. But for sure there are some other formulas for different summations. Is there something like a table ob how to solve summations? For example, what is when the index of the summation changes, like from n=0 to n=1 ? \begin{aligned} H[e^{j\omega}] & = \sum_{n=1}^{\infty} (\frac{1}{2} e^{-j\omega})^n = \frac{1}{1-0.5e^{-j\omega}}\\ \end{aligned} Because my second example is \begin{aligned} H_2(n) = h(n) - h(n-1) \end{aligned} –  madmax Apr 20 '11 at 13:19
Well, i think the solution has to be \begin{aligned} H[e^{j\omega}] & = \sum_{n=1}^{\infty} (r)^n = \frac{r}{1-r}\ \end{aligned} –  madmax Apr 20 '11 at 13:43
@madmax Check out en.wikipedia.org/wiki/Geometric_series where geometric series are explained in more detail. –  svenkatr Apr 20 '11 at 16:11