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I saw on a book to follwoing claim:
Given an Absolutely Summable Series $ \sum_{n = -\infty }^{\infty}\left | x\left [ n \right ] \right | \leqslant \infty $, Namely, $ l_1 $ series it is possible to show its DTFT (Discrete Time Fourier Transform) is continuous.
Where the DTFT is given by: $$ DTFT\left \{ x\left [ n \right ] \right \} = X\left ( {e}^{j \omega} \right ) = \sum_{n = -\infty}^{\infty} x\left [ n \right ] {e}^{-j \omega n} $$

My question is, how could that be proven?
Is it also differentiable?

Thank You.

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Please, Anyone? –  Drazick Nov 12 '12 at 18:38
    
Please, Anyone? –  Drazick Nov 23 '12 at 15:56
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