# Types of Convergence of the DTFT & Relation to Summability of x[n]

#### CAVision

Given a discrete time signal x[n] that has a DTFT (which exists in the mean square convergence or in the uniform convergence sense), how can we tell if the signal x[n] converges absolutely?

I know the following:

x[n] is absolutely summable <=> $X(e^{j \omega})$converges uniformly (i.e. the ROC of the Z-transform includes the unit circle)

x[n] is square summable <=>$X(e^{j\omega})$ converges in the mean-square sense (i.e. the ROC of the Z-transform does not include the unit circle)

Specifically, given x[n] with DTFT
$X(e^{j\omega}) = \frac{1 + 0.55e^{-j\omega} -0.2e^{-j2\omega} }{(1 + 0.8665e^{-j\omega} + 0.5625e^{-j2\omega})(1+2e^{-j\omega})}$
converges (uniformly or in mean-square). Is x[n] absolutely summable?

Thanks.

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