What we can say about $\lim_{n \to \infty} b_n$?

Question

Suppose for all $n \in \mathbb{N}$ $\lvert{a_n}\rvert \ge \lvert{b_n}\rvert$ and also $\lim_{n \to \infty} \lvert a_n \rvert = 0$ . What we can say about $\lim_{n \to \infty} b_n$ ? I think the answer is zero but I can't prove it using definition .

Answer 1

True, and you should be able to prove this via the sandwich (squeeze) theorem. I would use $\vert a_n \vert$ as one of your bounds, can you think of the other?

Answer 2

Suppose by contradiction that $b_n$ doesn’t tend to $0$ as $n \to \infty$.

It means that there is an $\epsilon > 0$ such that :

$$\exists (a_i)_{i \in \mathbb{N}}, a_i < a_{i +1}, \forall i, \mid b_{a_i}\mid >\epsilon$$

But since $a_n \to 0$, we can find an $N$ such that :

$$\forall n > N, \mid a_n\mid < \epsilon$$

And since : $\forall n, \mid a_n \mid \leq \mid b_n \mid$, then in particular we have :

$$\forall n > N, \epsilon >\mid a_n \mid \geq \mid b_n \mid$$

Hence it contradicts the existence of the sequence $(a_i)_{ i \in \mathbb{N}}$, so $b_n \to 0$.

Answer 3

One way is the next: since $|a_n| - |b_n| \geq 0$ for all $n = 1 , 2 , \ldots$, then $\lim_{n \to \infty} (|a_n| - |b_n|) \geq 0$. Using the basic properties about limits and the fact $\lim_{n \to \infty} |a_n| = 0$, the last inequality is equivalent to $- \lim_{n \to \infty} |b_n| \geq 0$. It implies automatically that $\lim_{n \to \infty} b_n = 0$.