If $(a_n)$ is an increasing sequence and $(b_n)$ is a decreasing sequence, with $a_n \leq b_n \forall n \in \mathbb N$. Prove that $\lim a_n \leq \lim b_n$

This is the closest thing I found on this site, but there the sequences are not given to be increasing and decreasing respectively.

My attempt:

$a_n \leq b_n \quad \forall n \in \mathbb N$

$\Rightarrow \sup_n a_n \leq (b_n) \quad \forall n \in \mathbb N$

So, $(a_n)$ is an increasing sequence that is bounded above.

Hence, by Monotone Convergence theorem, $\lim a_n=\sup a_n$

Again, $a_n \leq b_n \quad \forall n \in \mathbb N$

$\Rightarrow (a_n) \leq \inf_n b_n \quad \forall n \in \mathbb N$

So, $(b_n)$ is an decreasing sequence that is bounded below.

Hence, by Monotone Convergence theorem, $\lim b_n=\inf b_n$

Now, $\sup a_n \leq \inf b_n$ and so $\lim(a_n) \leq \lim(b_n)$

Is my proof correct? My confusion is with the $\sup a_n \leq b_n\> \forall n \in \mathbb N$ and $\inf b_n \>\forall n \in \mathbb{N}$ parts. Can I say these are true $\forall n \in \mathbb N$, or is it true for a fixed $n$?

  • $\begingroup$ You don't need them to be increasing or decreasing, that requirement is unnecessary. $\endgroup$
    – user223391
    May 3 '15 at 2:36
  • $\begingroup$ @avid19 but that's the given condition in my question, so I need to use it I think :) $\endgroup$
    – Diya
    May 3 '15 at 2:39
  • 1
    $\begingroup$ Increasing and decreasing are just used to guarantee the two limits exist. $\endgroup$
    – Brian Ding
    May 3 '15 at 2:44

It follows from induction that $a_n\leqslant b_1$ for all $n$, so $\lim_{n\to\infty}a_n=\sup_n a_n$ exists. Similarly, $b_n\geqslant a_1$ for all $n$, so $\lim_{n\to\infty}b_n = \inf_n b_n$ exists.

Since $a_n\leqslant b_n$ for all $n$, this inequality holds in the limit as $n\to\infty$.


Let $(x_n)$ be a convergent sequence (to $x\in \mathbb{R}$), with $x_n \geq 0$. Then $x\geq 0$.

Proof: Suppose (to reach a contradiction) that $x<0$. You have that, for every $\epsilon>0$, you can find $N>0$ such that $$ n>N\to x-\varepsilon<x_n<x+\varepsilon $$ Taking $\epsilon = |x|/2$, and that gives you $x_n<x+|x|/2 < 0$. (End of proof)

Having this, now suppose you have two convergent sequences $(a_n)$ and $(b_n)$, with $a_n\leq b_n$ for every $n$. Then, if $a_n\to a$ and $b_n\to b$, you have $(b_n-a_n) \to b-a$, and since $b_n-a_n\geq 0$, using the result above, you get $b-a\geq 0 \to b=\lim b_n\geq a=\lim a_n$. (You don't have to assume increasing/decreasing).


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