1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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 $a_n = b_n \times \frac{a_n}{b_n}$. $\endgroup$ Mar 24 '21 at 18:04
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – MrMath
    Mar 24 '21 at 20:10
  • 1
    $\begingroup$ @MrMath yep, I think that's right. $\endgroup$ Mar 24 '21 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.