# When limit superior diverges

• Mar 29th 2011, 03:28 PM
Connected
When limit superior diverges
Prove that $\displaystyle \lim\sup x_n=\infty\iff \forall M>0,\forall n\in\mathbb N,\exists k_0\ge n$ so that $\displaystyle x_k>M.$

I think is an easy problem, but I'm confused, the statement establishes that $\displaystyle x_k$ is bounded below, but not above.

How to prove this?
• Mar 29th 2011, 05:41 PM
Drexel28
Quote:

Originally Posted by Connected
Prove that $\displaystyle \lim\sup x_n=\infty\iff \forall M>0,\forall n\in\mathbb N,\exists k_0\ge n$ so that $\displaystyle x_k>M.$

I think is an easy problem, but I'm confused, the statement establishes that $\displaystyle x_k$ is bounded below, but not above.

How to prove this?

Which part are you having trouble with? Try firstly the only if statement. Then, you want to prove that $\displaystyle \limsup x_n=\infty$ but evidently by assumption you have that $\displaystyle \displaystyle \sup_{n\geqslant N}x_n\geqslant M$ for every $\displaystyle M\in\mathbb{R}^+$ so that $\displaystyle \displaystyle \limsup x_n=\lim_{N\to\infty}\sup_{n\geqslant N}x_n\geqslant M$ from where the conclusion follows since $\displaystyle M$ were arbitrary (intuitively you can take the limit as $\displaystyle M\to\infty$ of both sides of this last expression, but the left side is 'unaffected' by the limit since there is no $\displaystyle M$)