17 - Sandwiching

Sandwiching (first version)

Let $(a_n)$ and $(b_n)$ be real sequences with

$$0\le a_n\le b_n\text{ for all }n\ge 1$$

If $b_n\to 0$ as $n\to \infty$, then $a_n\to \infty$ as $n\to \infty$

Proof

Assume that $0\le a_n\le b_n$ for all $n$, and that $b_n\to 0$ as $n\to \infty$
Take $ϵ>0$
Since $b_n\to 0$, there exists $N$ such that

$$\text{if }n\ge N\text{ then }|b_n|<ϵ$$

Now if $n\ge N$ then $0\le a_n\le b_n<ϵ$, so $|a_n|<ϵ$
So $a_n\to 0$ as $n\to \infty$

Sandwiching

Let $(a_n),(b_n)\text{ and }(c_n)$ be real sequences with

$$a_n\le b_n\le c_n\text{ for all }n\ge 1$$

If

$$a_n\to L\text{ and }{c}_n\to L\text{ as }n\to \infty$$

Then

$$b_n\to L\text{ as }n\to \infty$$

Proof

Take $ϵ>0$
Since $a_n\to L$ as $n\to \infty$, there is $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-L|<ϵ$$

Since $c_n\to L$ as $n\to \infty$, there is $N_2$ such that

$$\text{if }n\ge N_2\text{ then }|c_n-L|<ϵ$$

Then for $n\ge \max \{N_1,N_2\}$ we have

$$L-ϵ\le a_n\le b_n\le c_n\le L+ϵ$$

Hence

$$|b_n-L|<ϵ$$

So $b_n\to L$ as $n\to \infty$