If $a_n \le b_n \le a_{n+1}$ prove that $\overline{\lim}_{n \to \infty} a_n=\overline{\lim}_{n \to \infty} b_n$

Question

If $a_n \le b_n \le a_{n+1}$ prove that $\overline{\lim}_{n \to \infty} a_n=\overline{\lim}_{n \to \infty} b_n=L\in\mathbb{R}$ or $\overline{\lim}_{n \to \infty} a_n=\overline{\lim}_{n \to \infty} b_n =\infty$

I can prove that if $\overline{\lim}_n a_n$ is something then exists a subsequence of $b_n$ with the same limit, but how do I prove it's the biggest sub-limit of $b_n$?

Answer 1

Let $\lim_{k}a_{n_{k}}=\limsup a_{n}$, then $a_{n_{k}}\leq b_{n_{k}}\leq a_{n_{k}+1}\leq a_{n_{k+1}}$, taking $k\rightarrow\infty$, we have $\limsup a_{n}\leq\lim_{k}b_{n_{k}}\leq\limsup a_{n}$, this shows that $\limsup b_{n}\geq\limsup a_{n}$.

Likewise, let $\limsup b_{n}=\lim_{l}b_{n_{l}}$, then $b_{n_{l}}\leq a_{n_{l}+1}\leq a_{n_{l+1}}$, taking $l\rightarrow\infty$, we have $\limsup b_{n}\leq\lim_{l}a_{n_{l}}\leq\limsup a_{n}$.

Note that $\{a_{n_{l}}\}$ is monotone.

Answer 2

HINT: if $a_n\le b_n$ then $\limsup a_n\le\limsup b_n$

Answer 3

Since you always have $a_n\leqslant a_{n+1}$, the sequence $(a_n)_{n\in\mathbb N}$ is increasing. Therefore, either it converges to some real number or $\lim_{n\to\infty}a_n=+\infty$. In any case, since $a_n\leqslant b_n\leqslant a_{n+1}$, $(b_n)_{n\in\mathbb N}$ will have the same limit.