Divergence to $+\infty$ and to $-\infty$

Let $(x_k)$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with $\mathbb{R}$ ordered as in Order on the reals and Complete ordered field (least-upper-bound property).

• $(x_k)$ diverges to $+\infty$, written $x_k \to +\infty$, when for every $M \in \mathbb{R}$ there is $K \in \mathbb{N}$ such that $x_k > M$ for all $k \ge K$.
• $(x_k)$ diverges to $-\infty$, written $x_k \to -\infty$, when for every $M \in \mathbb{R}$ there is $K \in \mathbb{N}$ such that $x_k < M$ for all $k \ge K$.

Equivalently, in the language of Sequences of reals: bounded, eventually, frequently, tails, subsequences: $x_k \to +\infty$ when the property $x_k > M$ holds eventually, for every real $M$.

• This is divergence, not convergence. The symbols $+\infty$ and $-\infty$ are not real numbers: $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property)) and contains no element larger than every element of itself. Nothing above claims that $(x_k)$ has a limit in the sense of Limits and Cauchy sequences of reals, and nothing above defines an object named $+\infty$. The whole phrase "$x_k \to +\infty$" is a single abbreviation for the displayed condition, exactly as "$(x_k)$ is Cauchy" is an abbreviation for a condition and not a claim that some object called a Cauchy value exists.

• A sequence diverging to $+\infty$ really does diverge. Suppose $x_k \to +\infty$. Given any real $M$, there is $K$ with $x_k > M$ for all $k \ge K$; in particular $x_K > M$, so no real $M$ satisfies $x_k \le M$ for all $k$. Since $x_k \le |x_k|$ always (Basic properties of the absolute value), a bound $|x_k| \le M$ valid for all $k$ would give $x_k \le M$ for all $k$, which has just been excluded, so no such $M$ exists either. Thus $(x_k)$ is unbounded, and an unbounded sequence cannot converge, since convergent sequences are bounded (Every convergent sequence is bounded). The same argument applies to $-\infty$. So the two notions never overlap: a sequence that diverges to $\pm\infty$ has no limit whatever.

• Consequently $\lim_k x_k$ is not written here. Many texts write $\lim_k x_k = +\infty$. This library does not, for the reason recorded in Conventions: $\sup \emptyset$, unbounded sets, and the extended reals about $\sup S = +\infty$: writing an equation whose right-hand side is not an element of $\mathbb{R}$ silently moves the discussion into the extended real line, a structure that is not a field, and every subsequent algebraic step then needs its own justification. In particular none of the rules of Algebra of limits: sums, scalar multiples, products and quotients may be applied to a divergence to $\pm\infty$; the familiar slogans "$\infty + \infty = \infty$" and "$\infty \cdot \infty = \infty$" are separate statements about this definition and would need separate proofs.

• Testing against naturals suffices. Since $\mathbb{R}$ is Archimedean (Every complete ordered field is Archimedean), every real $M$ is below some canonical natural $n$, so the condition "for every real $M$" may equivalently be read as "for every natural $n \ge 1$"; the two formulations of $x_k \to +\infty$ agree.

• Divergence to $+\infty$ is much stronger than divergence. A sequence alternating between $1$ and $-1$ diverges (FALSE: every bounded sequence converges) but goes to neither $+\infty$ nor $-\infty$, since it is bounded. Divergence is the negation of convergence; divergence to $+\infty$ is a positive statement about growth.