Definition of divergent series for series of real numbers and of complex numbers

Question

Divergence of a series of complex numbers is defined in the following way:

The series

$$\sum_{n \geq 0} a_n\,\,\,\,\, , a_n \in \mathbb{C}$$ diverges iff $$\lim_{N \to \infty} \bigg( \sum_{n=0}^{N} |a_n| \bigg)=+\infty$$

Does this definition hold as stated also if $a_n \in \mathbb{R}$?

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Or, in that case, do we consider the limit without the absolute values? That is

$$\lim_{N \to \infty} \bigg( \sum_{n=0}^{N} a_n \bigg)=\pm \infty$$

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Or maybe would it be the same thing to consider $\lim_{N \to \infty} \bigg( \sum_{n=0}^{N} a_n \bigg)$ or $\lim_{N \to \infty} \bigg( \sum_{n=0}^{N} |a_n| \bigg)$? In particular does the following hold?

$$\lim_{N \to \infty} \bigg( \sum_{n=0}^{N} |a_n| \bigg)=+ \infty \iff \lim_{N \to \infty} \bigg( \sum_{n=0}^{N} a_n \bigg)=\pm \infty\,\,\,\,\,\,\,\,\,\,\,\, a_n \in \mathbb{R}$$

Answer 1

The statement holds for all $a_n$ including when $a_n$ is an element of the reals. However, this is not the way the definition is normally stated. Generally, it is stated that a series is divergent if it is not convergent.

Answer 2

Your statement is false. Note that if

$$a_n=(-1)^n/n$$

The series will converge, but the absolute value of it will not. Thus, this is called a conditionally convergent series. If it satisfied your requirement, it would be absolutely convergent.

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The usual definition is as follows:

A sequence is said to converge to a limit $L$ if for all $\epsilon>0$, there exists $N$ such that for all $n>N$, we have

$$|L-s_n|<\epsilon$$

Particularly, we have

$$s_n=\sum_{k=1}^na_k$$

But if there does not exist an $N$ for any $\epsilon>0$, then it diverges.