# Properly Divergent Sequences

Using the definition in Bartle's Introduction to Real Analysis, I am trying to gain an intuitive understanding of limits that tend to infinity.

Given Definition:

Let ($x_n$) be a sequence of real numbers.

(i) We say that ($x_n$) tends to $\infty$, and write $lim(x_n) = +\infty$ , if for every $\alpha \in \Bbb R$ there exists a natural number $K(\alpha)$ such that if $n \ge K(\alpha)$, then $x_n > \alpha$.

(ii) We say that $(x_n)$ tends to $- \infty$, and write $lim(x_n) = - \infty$, if for every $\beta \in \Bbb R$ there exists a natural number $K(\beta)$ such that if $n \ge K(\alpha)$, then $x_n < \beta$.

My question is this: Can anyone put this into words or paraphrase their understanding of this concept? I'm having trouble grasping the concept of these kinds of limits with the formal definitions. Apologies if this is more of a 'soft' question.

As an example question, I know that the lim$(\sqrt{n^2+2} = \infty$ however I don't understand why this is so.

Essentially what the definition is saying is, if you claim that a sequence ${x_n}$ is truly 'approaching infinity', if I give you a really large number, say $100000000000000000000000$ you should be able to tell me that there is a point in this sequence of numbers $x_n$, where if you take all terms of the sequence after that point, they are all bigger than $100000000000000000000000$ .

Now you should not only be able to do this with $100000000000000000000000$, but literally with all positive numbers, that is numbers of ANY size, no matter how large.

So intuitively, this means that the sequence keeps getting larger and larger and never ceases to get larger and larger.

This is basically the same for when a sequence tends to $-\infty$, except the sequence gets increasingly large and negative.

You may think, okay so a sequence tending to $\infty$ intuitively means it gets larger and larger, so why don't we leave it at that? Point is, how do we actually know a sequence continues to get larger if we can't find terms in the sequence that exceed a number we have in mind? The answer is we can't know for sure, which is why we define a sequence to tend to $\infty$ only if we can do this, and for all positive numbers too.

EDIT: Also, just for completeness, we don't actually NEED to find a $K(\alpha)$ so that for all $n\geq K(\alpha)$ , $x_n \geq \alpha$ , we could use non-strict inequalities for the definition as well and it would be the same. That is, we could also just find a $K(\alpha)$ so that for all $n> K(\alpha)$ , $x_n > \alpha$.

• Thank you for your input. Where does the $\alpha$ and $K(\alpha)$ come in for this explanation. Are we setting $\alpha$ as the very large number?
– Iff
Dec 6, 2015 at 6:27
• Correct, $\alpha$ is the very large number, and $K(\alpha)$ is a number that represents the point in the sequence where if we take all terms after this point, the sequence is always larger than $\alpha$ Dec 6, 2015 at 6:30
• Got it, thank you so much!
– Iff
Dec 6, 2015 at 6:52

Let me use divergence to plus infinity ($+\infty$) as an example. The formal definition for a real sequence to diverge to plus infinity says that, given any real number $\alpha$, we can find some term of the sequence such that every term of the sequence after this term is greater than $\alpha$.

If $n \geq 1$, then, given any $\alpha \in \Bbb{R}$, we have $\sqrt{n^{2}+2} > \alpha$ if $n^{2} > \alpha^{2}-2$; hence, if $\alpha^{2} - 2 \leq 0$ then take $K := 1$, and if $\alpha^{2} - 2 > 0$ then take $K := \lceil \sqrt{\alpha^{2}-2} \rceil + 1$.

• Thank you, I appreciate your help!
– Iff
Dec 6, 2015 at 6:52