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?

  • 3
    $\begingroup$ Your proof is fine but I suggest using an upper bound instead of Cauchy property. $\lim \sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $\sum_{-N}^{N}|a_n| \leq M$ for all $n$ and the rest should be clear from this. $\endgroup$ – Kavi Rama Murthy Jan 3 at 6:15
  • 1
    $\begingroup$ Thanks, great suggestion! $\endgroup$ – zokomoko Jan 3 at 6:20

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