# On the convergence of a two-sided series

## 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?

• 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. – Kavi Rama Murthy Jan 3 at 6:15
• Thanks, great suggestion! – zokomoko Jan 3 at 6:20