MHB Can a sequence without a convergent subsequence have a limit of infinity?

[Fernando Revilla]

I quote a question from Yahoo! Answers

Let (xn)n be a sequence of positive real numbers that has no convergent subsequence. Prove that lim(n→∞) x of n=+∞. What if the xn are permitted to take both positive and negative values?

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

Each bounded sequence has a convergent subsequence, so $x_n$ has no convergent subsequence implies that $x_n$ is not bounded. In this case, $x_n\to +\infty$ if $x_n$ is a sequence of positive real numbers and $\left|x_n\right|\to +\infty$ if the $x_n$ are permitted to take both positive and negative values.

 

[chisigma]

Each bounded sequence has a convergent subsequence, so $x_n$ has no convergent subsequence implies that $x_n$ is not bounded. In this case, $x_n\to +\infty$ if $x_n$ is a sequence of positive real numbers and $\left|x_n\right|\to +\infty$...

The fact that the positive term sequence $x_{n}$ is unbounded doesn't mean that $\displaystyle \lim_{n \rightarrow \infty} x_{n} = + \infty$. As example You can consider the sequence $\displaystyle x_{n} = |\tan n|,\ n \ge 1$...

Kind regards

$\chi$ $\sigma$

 

[Fernando Revilla]

The fact that the positive term sequence $x_{n}$ is unbounded doesn't mean that $\displaystyle \lim_{n \rightarrow \infty} x_{n} = + \infty$. As example You can consider the sequence $\displaystyle x_{n} = |\tan n|,\ n \ge 1$...

Right, my fault, I was thinking about a monotic sequence.

 

[Evgeny.Makarov]

The negation of $\lim_{n\to\infty}x_n=\infty$ says that $x_n$ has a bounded subsequence and therefore a convergent sub-subsequence.