TITLE: Limit of a monotonically increasing sequence and decreasing sequence
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: Since $\lim_{n\to\infty}(b_n-a_n)=0$, there is an $N$ such that $|a_n-b_n|<1$ for all $n\ge N$. ($1$ is a number that I have just chosen for $\varepsilon$.) Since $b_n$ is decreasing, we have $a_n<b_n+1\le b_N+1$ for all $n\ge N$. Therefore $a$ is bounded from above, indeed by $\max\left\{a_0,a_1,\dots,a_{N-1},b_N+1\right\}$.
Therefore $a$ converges. Since $b_n = a_n + (b_n-a_n)$ and both $a$ and $b-a$ converge, $b$ also converges.