# Proof of 'sandwich theorem' for sequences

Please bear with me here and please try to read it all and spot any mistakes or errors as I'm trying to prove this result but I'm unsure of whether I have done it or not. THANK YOU.

Suppose we have the following statement $(a_n)\rightarrow \ell , (b_n)\rightarrow \ell$ and we have $$a_n\leq c_n\leq b_n$$ then $(c_n)\rightarrow \ell$. I think I have a proof which goes as follow ; $$a_n\leq c_n\leq b_n\leq \ \Rightarrow 0\leq c_n-a_n\leq b_n-a_n$$, as the terms are all larger than 0, taking the absolute value will not change any of the signs of the inequalities. So we have $$0\leq |c_n-a_n|\leq|b_n-a_n|.$$ Now consider $$|b_n-a_n|=|(b_n-\ell )+(\ell - a_n)|\leq |b_n-\ell |+|a_n - \ell | \text{ (by triangle inequality)} .$$ Using the definition of a sequence tending to a value, if $(a_n)\rightarrow \ell$ then $\exists N_1\in \mathbb{N} \text{ s.t} \ \forall n>N, |a_n-\ell |<\epsilon \ ,\forall \epsilon >0 .$ We do the same for $(b_n)$ but replacing $N_1$ with $N_2$ and using the same $\epsilon$ without loss of generality. So we can now say that $$|b_n-a_n|\leq |b_n-\ell |+|a_n - \ell |<2\epsilon .$$ So we have $$0\leq |c_n-a_n |\leq|b_n-a_n|<\epsilon _0, \text{ where } \epsilon _0=2\epsilon .$$ So we can conclude (using sandwich theorem for null sequences) that $(c_n-a_n)\rightarrow 0 \Rightarrow (c_n)\rightarrow (a_n)\Rightarrow (c_n)\rightarrow \ell$ since $(a_n)\rightarrow \ell . \ \ \ \square$

## 3 Answers

Seems OK, as far as you have sandwich theorem for null sequences.

Another way can be to choose an $\epsilon>0$ and notice that there is some $N\in \mathbb{N}$ such that for $n\geq N$, we have both $|a_n - \ell|<\epsilon$ and $|b_n-\ell|<\epsilon$. The last two inequalities imply $a_n-\epsilon<\ell<b_n+\epsilon$, and we also know that $a_n-\epsilon<c_n<b_n+\epsilon$.

So $|c_n-\ell |<2\epsilon$ whenever $n\geq N$. The last shows that $(c_n)$ converges to $\ell$

I would say directly that $$|c_n - \ell| \le |c_n - a_n|+|a_n-\ell|\\=(c_n-a_n)+|a_n-\ell|\\ \le(b_n-a_n)+|a_n-\ell|\\ =|b_n-a_n|+|a_n-\ell|\\ \le|b_n-\ell|+2|a_n-\ell|,$$ using (respectively) the triangle inequality; $c_n \ge a_n$; $c_n \le b_n$; $a_n \le b_n$; and the triangle inequality again. Then for any $\epsilon>0$, choose $N$ large enough that $|a_n-\ell|,|b_n-\ell| \le \epsilon/3$ for $n\ge N$; then $|c_n-\ell|\le \epsilon$ for $n\ge N$. Since $\epsilon$ was arbitrary, this proves $(c_n)\rightarrow\ell$.

For any $n$ such that $a_n,b_n \in (l-\epsilon,l+\epsilon),$ convexity shows $[a_n,b_n]\subset (l-\epsilon,l+\epsilon).$ Since $c_n \in [a_n,b_n],$ we have $c_n \in (l-\epsilon,l+\epsilon)$ for the same $n.$