Proof of 'sandwich theorem' for sequences

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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 $

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3 Answers

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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 $\ell$

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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 $(c_n)\rightarrow\ell$.

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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 $n.$

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