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

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?

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

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

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

$a_n\leq b_n\leq b_0$ 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, $lim_n(a_n-b_n)=lim_na_n-lim_nb_n=0$.

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

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.