Proof for Squeeze Theorem for Sequences

Question

Squeeze Theorem: If $a_n$, $b_n$ and $c_n$ are any three sequences with $a_n \lt b_n \lt c_n$ for any $n$ and $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = l$. Then $\lim_{n \to \infty} b_n = l$.

Proof: Since $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = l$ there are $N_a, N_c \in \mathbb{N}$ so that for any $n \gt N_a$ it holds that $\vert a_n - l \vert \lt \epsilon$ for any $\epsilon \gt 0$. Equaly, if $n \gt N_c$, $\vert c_n - l \vert \lt \epsilon$. We choose $N = max(N_a, N_c)$ so that, when $n \gt N$, both $\vert a_n - l \vert \lt \epsilon$ and $\vert c_n - l \vert \lt \epsilon$ are true.

Now, since $a_n \lt b_n \lt c_n$ it follows, that $\vert b_n - l \vert \le max(\vert a_n - l \vert , \vert c_n - l \vert)$. So, if $n \gt N$, then $\vert b_n - l \vert \lt \epsilon$ has to be true. Thus $\lim_{n \to \infty} b_n = l$. $\blacksquare$

Is this correct?

Answer 1

Nice attempt but you should be careful, it's really tricky to keep track of all the logical qualifiers at once! You seem to have the right idea, but you have mistakenly specified your $N_a$ as to hold for all $\epsilon>0$. This would imply that the sequence is eventually constant (equal to $l$), which may not be the case. What we can instead say when we have convergence is that $$\text{for any }\epsilon >0,\text{we can find some }N_a\text{ such that for}\ldots$$ This is different from $$\text{we can find some }N_a\text{ such that for any }\epsilon >0\ldots$$

As a silly example, for any positive number $\epsilon$, we can find some smaller positive number, say $\frac{\epsilon}{2}$. But we cannot find a positive number that is smaller than any other given positive number! (I'm obliged to mention that there are models of mathematics where we can, but that is far removed from the point and probably poor pedagogy here.)

I think you are on the right track, so I recommend that you try to "read the mathematical statement out loud" and see what happens to the precise meaning when you permute some of the quantifiers. You should notice a subtle change and after really taking care to follow the definition of convergence you should be able to fix it. Good luck!

Answer 2

Thanks for the answer @Cornell Alex Holmes. I think I knew the correct definition. But perhaps I didn't grasp it's subtleties and thus formulated in incorrectly. Here is my second try:

Squeeze Theorem: If $a_n$, $b_n$ and $c_n$ are any three sequences with $a_n \lt b_n \lt c_n$ for any $n$ and $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = l$. Then $\lim_{n \to \infty} b_n = l$.

Proof: Since $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = l$, for any $\epsilon \gt 0$, there are $N_a, N_c \in \mathbb{N}$ so that if $n \gt N_a$ it follows that $\vert a_n - l \vert \lt \epsilon$ . Equaly, if $n \gt N_c$ then $\vert c_n - l \vert \lt \epsilon$. We choose $N = max(N_a, N_c)$ so that, when $n \gt N$, both $\vert a_n - l \vert \lt \epsilon$ and $\vert c_n - l \vert \lt \epsilon$ are true.

Now, since $a_n \lt b_n \lt c_n$ it follows, that $\vert b_n - l \vert \le max(\vert a_n - l \vert , \vert c_n - l \vert)$. So, if $n \gt N$, then $\vert b_n - l \vert \lt \epsilon$ has to be true.

This means, that for for any $\epsilon \gt 0$ we can find a $N \in \mathbb{N}$ so that $n \gt N \implies \vert b_n - l \vert \lt \epsilon$. Thus $\lim_{n \to \infty} b_n = l$. $\blacksquare$