If $(a_n)\to A\neq 0$ and $(a_n b_n)\to AB$ then $(b_n)\to B$

Question

Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) \to A\neq 0$ and $(a_n b_n)\to AB$ then $(b_n)\to B$

I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that

$$|b_n - B| < \epsilon$$

I know that since $<a_n>\to A$ then there exists an $N_1\in \mathbb{N}$ such that for all $n > N_1$

$$|a_n - A| < \epsilon$$

Similarly, since $<a_n b_n>\to AB$ then there exists an $N_2\in\mathbb{N}$ such that for all $n > N_2$

$$|a_n b_n - AB| < \epsilon$$

Any help would be appreciated.

Background information:

Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$

Lemma 8.1 - If $a_n\neq 0$ for all $n\in\mathbb{N}$ and if $(a_n)$ converges to $A\neq 0$ then the sequence $(1/a_n)$ converges to $1/A$

Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.

I don't think this is complete can someone help me add details?

Answer 1

Your proof is correct but I'd recommend justifying that $a_n \neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression $$ b_n = \left( \frac{a_nb_n}{a_n} \right). $$ To justify that $a_n \neq 0$ for all $n$ sufficiently large, we will use that $A = \lim{a_n} \neq 0$. Because $a_n \to A$, there exists $N \in \mathbb{N}$ such that $$ |a_n - A| < \epsilon = \frac{|A|}{2} $$ for all $n \geq N$. Therefore, for all such $n$, \begin{align*} |A| - |a_n| \leq \left\vert a_n - A \right\vert< \frac{|A|}{2} \end{align*} whence $$ 0 < \frac{|A|}{2} < |a_n|, \quad \forall n\geq N. $$ This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence $$ \frac{b_n}{a_n} $$ is well defined for all $n \geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that \begin{align*} \lim{b_n} = \lim\left(\frac{a_n b_n}{a_n} \right) = \frac{\lim(a_nb_n)}{\lim{a_n}} = \frac{AB}{A} = B. \end{align*}

Answer 2

Note that if $a_n \to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| \le L$.

Suppose $b_n \to b$ then $|a_nb_n-ab| \le |a_n b_n -a_n b| + |a_n b - ab| \le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n \to ab$.

Suppose $a \neq 0$. Since $a_n \to a$, we see that $|a_n| \ge {1\over 2} |a|$ for $n$ sufficiently large (and hence non zero). Then $|{1 \over a} - {1\over a_n}| = {|a_n-a| \over |a a_n|} \le {2 \over |a|^2} |a_n-a|$ and hence ${1 \over a_n} \to {1 \over a}$.

Hence if $a_nb_n \to ab$ and $a_n \to a \neq 0$ then $a_n b_n {1 \over a_n} = b_n \to ab {1 \over a} = b$.