0
$\begingroup$

I know that, if a sequence has no upper bound, it doesn't necessarily mean that it is going off to infinity. However, I don't know how to prove that this isn't the case when my sequence is increasing.

I know that $u_n \le u_{n+1}$, and I want to say that $u$ has to be going to infinity, and therefore there's a point at which $u$ is positive, but I can't use that definition. Is there a more precise definition for a sequence with no upper bound to show that the title is true, or am I going completely the wrong way? Thanks.

$\endgroup$
2
  • 1
    $\begingroup$ If a sequence has no upper bound, then $0$ is not an upper bound, so there is a sequence term with $u_n > 0$. $\endgroup$
    – user296602
    Apr 16 '18 at 20:24
  • $\begingroup$ The statement that sounds like what you are saying is that a sequence without a maximum doesn't necessarily go to infinity. $\endgroup$
    – user551819
    Apr 16 '18 at 20:30
1
$\begingroup$

Pick your favorite real number $L$.

$L$ is not an upper bound of the sequence, so there exists an index $N$ with $u_N > L$.

Since $\{u_n\}$ is increasing, $n \ge N \implies u_n \ge u_N \implies u_n > L$.

That is, for every $L \in \mathbb R$ there exists $N \in \mathbb N$ with the property that $$n \ge N \implies u_n > L.$$

This is the definition of $u_n \to \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.