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For instance consider the sequence $\{1,0,2,0,3,0,4,0,..\}$ Intuitively we know that the sequence converges to $\infty$ but how do we check that rigorously. If I imitate the formal definition of convergence then I believe that we can at best come up with something like this:

$(x_n)\to\infty$ if for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for $n\geq N$ we have $x_n>\epsilon.$

Now this definiton does help us in proving the convergence of some sequences such as $x_n=\sqrt{n}$ because in this case we can let $N\geq \epsilon^2.$ However this definition fails to show that the aforementioned sequence $\{1,0,2,0,3,0,4,0,..\}$ converges to $\infty$. I am thus guessing that "there exists" a better definition out there. So please suggest me some references or maybe provide me with a definition that is able to take care of convergence to $\infty$ in general.

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    The sequence $(1, 0, 2, 0, \dots)$ does not converge to $\infty$. – 2017-01-08
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    The sequence $1,0,2,0,3,0,4,0,\ldots$ is not something that I would describe as "converging to infinity," just as the sequence $1,1/2,1,1/3,1,1/4,1,1/5,1,\ldots$ does not converge to zero. However, its $\limsup$ is $\infty.$ – 2017-01-08
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    If your sequence $(1,0,2,0,3,0,4,0,...)$ converged to infinity then so would every subsequence, particularly the subsequence $(0,0,0,0,...)$. Which is absurd by anyone's definition of "convergence to infinity". – 2017-01-08
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    There is, afaik, nothing like "convergence to $\inf$". If a sequence tends to go to infitiy, it diverges, just as it would, as if it would be jumping (just like this sequence is) – 2017-01-08
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    converging to $+\infty$: there is no subsequence bounded from above, converging to $-\infty$: there is no subsequence bounded from below. – 2017-01-08
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    You are using an improper language: the sequence $1,0,2,0,3,0,4,0,...$ is not "converging to $\infty$" because it has a subsequence $0,0,0,0,...$ which does not approach to $\infty$. I think that you are looking for the condition of being an unbounded sequence: this has a simple definition $$x_n \mbox{ is unbounded (above)} \ \ \ \Leftrightarrow \ \ \ \forall \varepsilon > 0 \ \exists N : x_N> \varepsilon$$ – 2017-01-08
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    Your sequence doesnt converge at infinity, instead it have two **cluster points** in $\bar{\Bbb R}$, one in $\infty$ and the other in zero. – 2017-01-08

4 Answers 4

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The general idea behind limits converging to $\infty$ are:

  • There is a topological notion of limit in terms of open sets or open neighborhoods rather than in terms of distance
  • $+\infty$ and $-\infty$ are best understood as points in the extended real numbers

The definition you cited is, in fact, equivalent to what it means for a real-valued sequence to converge to $+\infty$ in the extended real numbers.

The problem is that the sequence you consider doesn't converge. In the extended real numbers, it is a divergent sequence with two limit points: $0$ and $+\infty$. In that regard, it's comparable to the sequence $0,1,0,1,0,1,\ldots$.

Without more examples or attempts at elaboration, I'm not sure what idea you have that you have incorrectly given the name "converging to $\infty$". One possibility is that you simply have in mind the idea of a sequence being unbounded above. This is a sequence having $+\infty$ as a limit point, or alternatively, something satisfying the property

For all real $M$, there exists some $n$ such that $x_n > M$

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The usual definition is that for every real number $x$ there is a positive integer $N$ such that $a_m>x$ for all $m\geq N$.

It is very similar to the usual definition of convergence, instead that we ask that all numbers be sufficiently "big" after a point. (instead of asking all numbers be sufficiently "close" to the limit after a point).

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    I'd find this answer a lot better if it would somehow take into account that this very definition is essentially already given in OP. As is, it's a bit confusing. – 2017-01-08
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We say that the sequence $(x_n)$ diverges to infinity. This is therefore different from a sequence like $\left((-1)^n\right)$, which diverges but not to infinity.

The formal definition you give for divergence to infinity is essentially correct.

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    The question is asking for a formal definition. What is the definition of divergence to infinity, and can you add it to your answer? – 2017-01-08
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    It is also common to say it *converges* to infinity. When set up correctly It *is* a simple convergence. – 2017-01-08
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Your definition is fine (even if "converge to infinity" is a sub-optimal choice of words since convergence to a finite limit implies many useful properties that tending to infinity does not), it is the sequence that does not converge. If a sequence converges (or tends to infinity), then every subsequence should converge to the same limit (or tend to infinity). This does not hold for the subsequence $\{0,0,0,...\}$ (which does not tend to infinity).

The limit superior $\left(\limsup_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} \sup_{m \ge n} x_n\right)$ is infinite, but that is not sufficient for a claim of "tends to infinity". As Will R pointed out in a comment, the sequence $\{ 1,1,1,{1\over 2},1,{1\over 3},1,{1\over 4},...\}$ does not converge to zero despite its limit inferior being zero. Another (necessary but insufficient for a claim of "tends to infinity") property of the sequence is that it is not bounded from above $\left( \forall M \ \exists n \mid x_n \gt M \right)$ (as mentioned by Hurkyl)

A possibly useful (although not perfect) way of thinking about the definition of tending to infinity $\left( \lim_{n \rightarrow \infty} x_n = \infty \Leftrightarrow \forall k \ \exists N \mid n \gt N \Rightarrow x_n \gt k\right)$ is "every subsequence exceeds all bounds".

Sources: currently studying Computer Science