Dividing Divergent and Convergent Sequences

Let $$a_n$$ be a diverging sequence and $$b_n$$ be a converging sequence. With that in mind does $$\frac{a_n}{b_n}$$ always diverge?

If so, what is the proof for this?

• One of the intermediate results in Real Analysis is that if $\langle r_n\rangle$ and $\langle s_n\rangle$ are two convergent sequences, then the sequence $$\langle {r_n} \times {s_n}\rangle$$ is also a convergent sequence. You can regard $a_n = b_n \times \frac{a_n}{b_n}$. Mar 24 '21 at 18:04

Let $$b_n$$ converge to $$l_1$$, and suppose $$\frac{a_n}{b_n}$$ converges to $$l_2$$. Then by the Algebraic Limit Theorem we have$$\lim_{n\to \infty}b_n\cdot \frac{a_n}{b_n} = l_1l_2.$$ But $$b_n \cdot \frac{a_n}{b_n} =a_n,$$ which is divergent, thus giving us a contradiction. Therefore $$\frac{a_n}{b_n}$$ is divergent.
• Thank you, following on from this if the opposite were to happen $a_n$ is convergent and $b_n$ is divergent will $\frac{a_n}{b_n}$ always either be divergent or converge to zero? Mar 24 '21 at 20:10