Question: Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences with $a_n \Rightarrow L$ and $b_n \Rightarrow M$ as $n \Rightarrow \infty$.

Prove that if $M \neq 0$, then $\frac{a_n}{b_n} \Rightarrow \frac{L}{M}$

My solution:


Let $B := max(|M - \epsilon|, |M + \epsilon|, |b_n|, \text{for } n > N)$

(1) $\exists L \in R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in N$, if $n > N_1$, then $|a_n - L| < B\frac{\epsilon}{2}$

(2) $\exists M \in R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in N$, if $n > N_2$, then $|b_n - M| < \frac{\epsilon |M|B}{2(|L|+1)}$

Let $\epsilon > 0$ be arbitrary

Choose N = $max(N_1, N_2) > 0$

Suppose $n > N$, then

$$|\frac{a_n}{b_n} - \frac{L}{M}| = |\frac{Ma_n - Lb_n}{b_nM}|$$

$$= |\frac{Ma_n + LM - LM - Lb_n}{b_nM}| = |\frac{M(a_n - L) + L(M-b_n)}{b_nM}|$$

$$\leq \frac{|M(a_n - L)| + |L(b_n - M)|}{|b_nM|} \text{ By triangle inequality}$$

$$< \frac{|(a_n - L)|}{|b_n|} + \frac{|(|L|+1)(b_n - M)|}{|b_nM|}$$

$$< B\frac{\epsilon}{2B} + \frac{\epsilon|M|(|L|+1)B}{2(|L|+1)B|M|}$$

$$= \epsilon$$

  • $\begingroup$ $|b_n|$ is bounded below by $|M|/2$ for almost all $n$. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 8:50
  • $\begingroup$ Okay, whether $L\neq 0$ or $L=0$, just make a little improvement by saying that $$|b_n - M| < \frac{\epsilon |M|B}{2(|L|+1)}$$ and with this we get $$< B\frac{\epsilon}{2B} + \frac{\epsilon|M||L|B}{2(|L|+1)B|M|}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ since $\frac{|L|}{|L|+1}<1$. $\endgroup$ – Juniven Mar 24 '17 at 8:52
  • $\begingroup$ Two MathJax/LaTeX hints: you can use \to instead of \Rightarrow for 'tends to' or 'converges to'; you can also use \left & \right modifiers to brackets or modulus vertical bars to make them automatically adjust their size to the contents — see \left(\frac 23\right) $\left(\frac 23\right)$ compared to (\frac 23) $(\frac 23)$. $\endgroup$ – CiaPan Mar 24 '17 at 8:54

If $L=0$ and $b_n \to M$ so $\tfrac{|M|}{2} \le \left| b_n \right| \le \tfrac{3|M|}{2}$ for $n$ sufficiently large $(*)$, then: $$\left| \frac{a_n}{b_n} - \frac{L}{M} \right| = \left| \frac{a_n}{b_n}\right| \le 2\frac{a_n}{|M|}$$ Now since $a_n \to 0$, pick $N$ to get $|a_n| < \tfrac{|M|\varepsilon}{2}$ and such that $(*)$ holds to finish.

Usually splitting into two cases is avoided and you could use ΘΣΦGenSan's suggestion from the comments. Since for $L \ne 0$, $\tfrac{1}{|L|}<\tfrac{1}{|L|+1}$, you can write for any $L$: $$|b_n - M| < \frac{\epsilon |M|B}{2\left(|L|+1\right)}$$ Note that the right-hand side works as an upper bound (compare it to yours) and the denominator can never become $0$, not even when $L=0$.

  • $\begingroup$ Maybe you want to delete the fraction $\frac{\epsilon |M|B}{2|L|}$? $\endgroup$ – Juniven Mar 24 '17 at 9:03
  • $\begingroup$ @ΘΣΦGenSan I wanted to stress that the new upper bound is bigger than the one OP originally took, but you are right that notation was sloppy (if $L=0$); I edited, thanks. $\endgroup$ – StackTD Mar 24 '17 at 9:06

It is easier this way. First show that if the sequence $a_n\rightarrow a$ and the sequence $b_n\rightarrow b$ then the sequence $a_nb_n\rightarrow ab$.

Now show that if $c_n\rightarrow c$ and $c\neq 0$ then $\frac{1}{c_n}\rightarrow \frac{1}{c}$. Then use both these facts to conclude your answer.


Wow, no offense, but your proof seems to me so cumbersome! We actually can do it in one stroke neatly.

To begin with, I suppose you mean there is some $N_{1}$ such that $b_{n} \neq 0$ for all $n \geq N_{1}$ (to ensure that $a_{n}/b_{n}$ is at least eventually meaningful.). Note that $$ |\frac{a_{n}}{b_{n}} - \frac{L}{M}| = | \frac{Ma_{n} - Lb_{n}}{Mb_{n}}| = \frac{|Ma_{n} - Lb_{n}|}{|M||b_{n}|} = \frac{|M(a_{n} - L) - L(b_{n} - M)|}{|M||b_{n}|} $$ for all $n \geq N_{1}$. Since $b_{n} \to M$, there is some $N_{2}$ such that $||b_{n}| - |M|| \leq |b_{n}-M| < |M|/2$ for all $n \geq N_{2}$; so $|b_{n}| > |M|/2$ for all $n \geq N_{2}$. Then
$$ |\frac{a_{n}}{b_{n}} - \frac{L}{M}| < \frac{2|M(a_{n} - L) - L(b_{n} - M)|}{M^{2}} \leq \frac{2|a_{n}-L|}{|M|} + \frac{2|L||b_{n}-M|}{M^{2}} $$ for all $n \geq \max \{N_{1},N_{2} \}$. Now again by the convergence assumption, given any $\varepsilon > 0$, there are some $N_{3},N_{4}$ such that $|a_{n}-L| < |M|\varepsilon/4$ for all $n \geq N_{3}$ and $|b_{n}-M| < M^{2}\varepsilon/4|L|$ for all $n \geq N_{4}$ (if $L = 0$ only $N_{3}$ is needed to bound away $|a_{n}-L|$). All in all, if $L = 0$ , taking $N := \max \{N_{1}, N_{2},N_{3} \}$ suffices; if $L \neq 0$, taking $N := \max \{N_{1},N_{2},N_{3},N_{4} \}$ suffices.

  • $\begingroup$ Yes it is quite long but its the exact definition in my textbook. Thx for your shorter way though $\endgroup$ – user349557 Mar 24 '17 at 18:37
  • $\begingroup$ ? You mean that proof is from book? Oh, u know, I would say my way above is more elegant :); but this is not just because that is from me :). $\endgroup$ – Megadeth Mar 24 '17 at 18:42
  • $\begingroup$ The proof is a exercise question in the book that i dont have the solution to, but the book took up a easier proof of $\{a_n +b_n\} \to (a + b)$. Which is much easier than any of the other properties but they did it this way. $\endgroup$ – user349557 Mar 24 '17 at 18:46

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