[Math] Limit of a monotonically increasing sequence and decreasing sequence

real-analysissequences-and-series

If a sequence ($a_n$) is monotonically increasing, and ($b_n$) is a decreasing sequence, with $\lim_{n\to\infty}\,(b_n-a_n)=0$, show that $\lim a_n$ and $\lim b_n$ both exist, and that $\lim a_n=\lim b_n$.

My attempt:

To show that the limits of both sequences exist, I think I should be using the Monotone Convergence Theorem (MCT). For that I would need to show that the sequences are bounded.

($a_n$) is increasing, and so it should be bounded below. ($b_n$) is decreasing, so it should be bounded above. The challenge here is to show that ($a_n$) can be bounded above and ($b_n$) can be bounded below. This should utilise the third condition, from which I get:

$$\begin{align*}
& \lim_{n\to\infty}\,(b_n-a_n)=0 \\[3pt]
\iff & \forall\varepsilon>0,\ \exists N\in \mathbb{N} \text{ s.t. } \forall n\geq N,\ |{b_n-a_n}|<\varepsilon
\end{align*}$$

I then tried using the triangle inequality:
$$ |b_n|-|a_n|\leq|b_n-a_n|<\varepsilon$$

but I'm not sure where to go from here.

Best Answer

The leftmost part of the inequality $|b_n|-|a_n|\leq|b_n-a_n|<\epsilon$ isn't needed in the proof.

Claim 1     $(a_n)$ is bounded above.

Proof    Let $\epsilon>0$. From $|b_{n+1}-a_{n+1}|<\epsilon$, $a_{n+1}<b_{n+1}+\epsilon$. Use the monotonicity of $(a_n)$ and $(b_n)$. We have $$a_1 \le \dots \le a_n\le a_{n+1} < b_{n+1}+\epsilon \le b_n+\epsilon \le \dots \le b_1 + \epsilon.$$ Since the choice of $n$ in the above inequality is arbitrary, we have $a_n < b_1 + \epsilon$ for all $n \in \Bbb N$. Therefore, $(a_n)$ is bounded above by $b_1 + \epsilon$.

Similarly, we have another claim.

Claim 2     $(b_n)$ is bounded below.

Now, recall that $(a_n)$ and $(b_n)$ are increasing and decreasing sequences respectively, and apply MCT to $(a_n)$ and $(b_n)$ to establish the existence of $\lim a_n$ and $\lim b_n$. Finally, use $\lim\limits_{n\to+\infty}(b_n-a_n)=0$ to conclude that $\lim a_n = \lim b_n$.

Sorry for using others' ideas in my solution. I would like to draw a commutative diagrams in the comments, but the system forbids me from posting comments with two or more @ characters, so I can't post the following diagram in a comment. Hoping that others can benefit from his answer at the first glance, I draw this diagram for fun.

A graphical explanation to DonAntonio's answer

$\require{AMScd}$ \begin{CD} @. a_n \\ @. @AA \vdots A \\ @. a_{N+1} \\ @. @AA (a_n)\uparrow A \\ @. a_N \\ \text{Suppose }a_N > b_K. \\ b_K @. \\ @V(b_n)\downarrow VV @.\\ b_{K+1} @. \\ @V \vdots VV @.\\ b_n @. \end{CD} It's clear from the diagram that we have to take $n \ge \max\{K,N\}$. But $\lim\limits_{n\to+\infty}(b_n-a_n)=0$, contradiction.

Related Question