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I know the following:

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

x[n] is square summable <=>[itex] X(e^{j\omega}) [/itex] 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

[itex] 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})} [/itex]

converges (uniformly or in mean-square). Is x[n] absolutely summable?

Thanks.