showing $\underset{n\rightarrow\infty}{\lim}a_{n}=\underset{n\rightarrow\infty}{\lim}b_{n}$

Question

Given the following conditions: $$\forall n\in \mathbb{N}: a_n \leq a_{n+1} ; b_n \geq b_{n+1} $$ $$\forall n\in \mathbb{N}: a_n \leq b_{n} $$ $$\underset{n\rightarrow\infty}{\lim}a_{n}-b_{n} = 0$$ I am trying to prove that $\underset{n\rightarrow\infty}{\lim}a_{n}=\underset{n\rightarrow\infty}{\lim}b_{n}$

I tried to use the Epsilon-approach, squeeze theorem and tried to bound those sequences and because both are monotonic to say the converge, but still couldn't manage to find any bound on those sequences. Any suggestions?

Answer 1

$$a_1 \le a_2 \le \ldots a_n \leq b_n \le \ldots b_1$$

The sequence $a_i$ is bounded above by $b_1$. Hence by monotone convergence theorem, $\lim_{n \to \infty} a_n$ exists.

Similarly, $\forall i, b_i \ge a_1$.

Answer 2

$a_n\leq b_n\leq b_0$ implies that $a_n$ converges since it is an increasing bounded sequence. $b_n\geq a_n\geq b_0$ implies that $b_n$ converges since it is a decreasing bounded sequence, $lim_n(a_n-b_n)=lim_na_n-lim_nb_n=0$.

Answer 3

$$a_n\leq a_{n+1}\leq b_{n+1}\leq b_n\implies 0\leq a_{n+1}-a_n\leq b_n-a_n\tag 1$$

Using $\lim\limits_{n\to\infty}b_n-a_n=0$, apply the squeeze theorem on $(1)$ to conclude that $\{a_{n+1}-a_n\}_{n\geq 1}$ converges to $0$ as $n\to\infty$, which is equivalent to the convergence of $\{a_n\}_{n\geq 1}$ to a finite limit as $n\to\infty$

In a similar vein, you can show that $\{b_{n+1}-b_n\}_{n\geq 1}$ converges to $0$, i.e., $\{b_n\}_{n\geq 1}$ converges to a finite limit as $n\to\infty$

Since $a_n-b_n\to 0$ and each of the individual limits exist finitely, we arrive at the required result.