2
$\begingroup$

PROPOSITION *Let $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ be convergent sequences, with $$\lim_{n \to \infty}a_n = a , \: \lim_{n \to \infty}b_n = b$$ then $$\lim_{n \to \infty}\frac{a_n}{b_n} = \frac{a}{b}$$ PROOF

Let $\varepsilon > 0$ be given and choose$K \in \mathbb{R}$ such that $\forall n \in \mathbb{N}$ we have that $\frac{1}{|b_n|} \leq K$. We can do this since the sequence is convergent and therefore bounded, so clearly the sequence $\frac{1}{|b_n|}$ will too be bounded.

Since $\lim_{n \to \infty}a_n = a$ then, $\exists \: N_1 \in \mathbb{N}$ such that $\forall n \geq N_1$ we have that $|a_n - a| \leq \frac{\varepsilon}{2K}$. Also since $\lim_{n \to \infty}b_n = b$ then, $\exists \: N_2 \in \mathbb{N}$ such that $\forall n \geq N_2$ we have that $|b_n - b| \leq \frac{|b|}{|a|} \frac{\varepsilon}{2K}$.

We have that \begin{align*}\left|\frac{a_n}{b_n} - \frac{a}{b}\right| &= \left|\frac{a_nb - ab_n}{b_nb}\right| \\ &= \left|\frac{a_nb - ba - ab_n + ba}{b_nb}\right| \\ &=\left|\frac{b(a_n - a)- a(b_n -b)}{b_nb}\right| \\ &= \left|\frac{(a_n-a)}{b_n} - \frac{a(b_n-b)}{b_nb}\right|\end{align*} Via the triangle inequality we have that \begin{align*}\left|\frac{a_n}{b_n} - \frac{a}{b}\right| =\left|\frac{(a_n-a)}{b_n} - \frac{a(b_n-b)}{b_nb}\right| \leq \left|\frac{a_n-a}{b_n}\right| + \left|\frac{a(b_n-b)}{b_nb}\right| = \frac{|a_n -a|}{|b_n|} + \frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|}\end{align*}

Since $|a_n - a| \leq \frac{\varepsilon}{2K}$ then we have that $$\frac{|a_n -a|}{|b_n|} \leq \frac{\varepsilon}{|b_n|2K}$$ Also since $|b_n - b| \leq \frac{|b|}{|a|}\frac{\varepsilon}{2K}$ then $$\frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{|b|}{|a|}\frac{\varepsilon}{2K}\cdot\left(\frac{|a|}{|b|}\frac{1}{|b_n|}\right) = \frac{\varepsilon}{2K|b_n|}$$

Since $\frac{1}{|b_n|} \leq K$ then $$\frac{\varepsilon}{2K|b_n|} \leq K\cdot\frac{\varepsilon}{2K} = \frac{\varepsilon}{2} \implies \frac{|a_n - a|}{|b_n|} \leq \frac{\varepsilon}{2}$$ \begin{align*} \frac{\varepsilon}{2K|b_n|} \leq \frac{\varepsilon}{2K}\cdot K = \frac{\varepsilon}{2} \implies\frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{\varepsilon}{2}\end{align*}

So we can deduce that $$\frac{|a_n -a|}{|b_n|} + \frac{|a|}{|b|}\frac{|b_n-b|}{|b_n|} \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

So then $$\left|\frac{a_n}{b_n} - \frac{a}{b}\right| \leq \varepsilon$$

This is true for $N \in \mathbb{N}$ when $n \geq N$ where $N = max\{N_1, N_2\}$, our choice of $\varepsilon$ was arbitrary so holds for all $\varepsilon > 0$. So indeed we have that $\lim_{bn \to \infty}\frac{a_n}{b_n} = \frac{a}{b}$.

$\endgroup$
3
  • 1
    $\begingroup$ $b_n$ should not tend to $0$. $\endgroup$ Apr 5, 2021 at 19:40
  • $\begingroup$ Thanks, I was aware of this, is the proof otherwise ok however? $\endgroup$
    – mosthigh
    Apr 5, 2021 at 19:41
  • $\begingroup$ @mosthigh It can’t be OK as nowhere you use the required hypothesis $b \neq 0$. $\endgroup$ Apr 5, 2021 at 19:43

1 Answer 1

1
$\begingroup$

It seems that in « your proof » you use at several steps the inequality

$$\frac{1}{\vert b_n \vert} \le K$$ which is not correct.

The proposition is wrong by the way if $b =0$. You have to add the hypothesis $b \neq 0$ which implies as $\{b_n\}$ is supposed to be a convergent sequence that

$$\vert b_n \vert \ge \frac{\vert b \vert}{2}\gt 0$$ for $n$ large enough and then correct your proof accordingly.

$\endgroup$
3
  • $\begingroup$ Thank you, I felt as though I rushed that part. $\endgroup$
    – mosthigh
    Apr 5, 2021 at 20:30
  • $\begingroup$ I have edited my original post, would you be so kind as to see if I have corrected it? $\endgroup$
    – mosthigh
    Apr 5, 2021 at 21:19
  • $\begingroup$ Your update is still not correct. You have in fact two constants. One, your original $K$ such that $\vert b_n \vert \le K$ valid for all $n$ because $\{b_n\}$ converges. And a second one $K^\prime$ such that $\frac{1}{\vert b_n \vert} \le K^\prime$ for $n$ large enough that exists because $\{b_n\}$ is supposed to converge to a non zero value. You need to explain that in your proof. $\endgroup$ Apr 6, 2021 at 5:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .