Background
Let $\left\{ {{a_n}} \right\}_{ - \infty }^\infty $ be a two sided sequence (is there a more proper term?) of complex numbers.
As far as I know (please correct me if I am wrong) we say that $$\sum\limits_{n = - \infty }^\infty {{a_n}} $$ converges if and only if the following two series converge: $$\sum\limits_{n = - \infty }^{ - 1} {{a_n}} ,\sum\limits_{n = 0}^\infty {{a_n}} $$
We know that the existence of $\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = - N}^N {{a_n}} $ does not imply the convergence of $\sum\limits_{n = - \infty }^\infty {{a_n}} $
since for example $\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = - N}^N n $ exists and the limit is zero (it is in fact the zero sequence, which limit is also zero) but the series $\sum\limits_{n = 0}^\infty n $ does not converge, therefore $\sum\limits_{n = - \infty }^\infty {{a_n}} $ diverges.
My question is the following:
(1) Let's say $\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = - N}^N {\left| {{a_n}} \right|} $ exists. Does that imply that $\sum\limits_{n = - \infty }^\infty {{a_n}} $ converges? converges absolutely? I think so and I'll try to prove it.
My attempt:
Let $\varepsilon > 0$.
Since the limit $\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = - N}^N {\left| {{a_n}} \right|} $ exists, by the Cauchy criterion we have a natural number ${N_0} $ such that for all $N > {N_0}$ and for all natural $p$ the following takes hold:
$\sum\limits_{n = - N + p}^{N + p} {\left| {{a_n}} \right|} - \sum\limits_{n = - N}^N {\left| {{a_n}} \right|} = \sum\limits_{ - \left( {N + p} \right)}^{ - \left( {N + 1} \right)} {\left| {{a_n}} \right|} + \sum\limits_{N + 1}^{N + p} {\left| {{a_n}} \right|} < \varepsilon $
But that implies
(1) $\sum\limits_{N + 1}^{N + p} {\left| {{a_n}} \right|} < \varepsilon $ so by Cauchy's criterion the series $\sum\limits_{n = 0}^\infty {{a_n}} $ converges absolutely.
(2) $\sum\limits_{-(N + p)}^{-(N + 1)} {\left| {{a_n}} \right|} < \varepsilon $ so by Cauchy's criterion the series $\sum\limits_{n = -\infty}^{-1} {{a_n}} $ converges absolutely.
Thus $\sum\limits_{n = - \infty }^\infty {{a_n}} $ converge absolutely and therefore converges.
Am I correct?