1
$\begingroup$

My solution:

Case 1: $b \neq 0$

WTS:

(1) $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in \mathbb N$,

if $n > N_1$, then $|a_n - a| < \frac{\epsilon}{2|b|}$

(2) $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$,

$$\text{Let } M := max(|a-\epsilon|, |a + \epsilon|, |a_n| \text{ for n} < N)$$

if $n > N_2$, then $|b_n - b| < \frac{\epsilon}{2M}$

Let $\epsilon > 0$ be arbitrary

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

Suppose $n > N$, then

$$|a_nb_n - ab| = |a_nb_n - a_nb + a_nb - ab|$$

$$=|a_n (b_n - b) + b(a_n - a)| \text{ By algebra}$$

$$\leq |a_n||b_n - b| + |b| |a_n - a| \text{ By triangle inequality}$$

$$< M\frac{\epsilon}{2M} + |b|\frac{\epsilon}{2|b|}$$

$$= \epsilon/2 + \epsilon/2 = \epsilon$$

$\endgroup$
  • $\begingroup$ You don't need to change the definitions. What you have to do is sgow that if $(a_n)$ converges, then it is bounded. Then you will get an inequality that looks like $|a_nb_n - ab| < c\epsilon$, where $c$ is some real number that does not depend on $n$. Then you will see that you need to be more careful when you choose $N$ (for instance choosing $N_1, N_2$ that work, not for $\epsilon$, but for...) $\endgroup$ – Max Mar 22 '17 at 19:49
  • $\begingroup$ What I'm trying to say is that how would I use $a_n$ in a way that is valid. For instance if I let $|a_n - a| < \frac{\epsilon}{|a_n|}$, and $|b_n - b| < \frac{\epsilon}{|b|}$. This would give me $\epsilon$ as wanted but I'm not allowed to use $a_n$ in the denominator. I'm asking whats the most simplest way to use $a_n$ in a way its allowed. $\endgroup$ – Tinler Mar 22 '17 at 19:52
  • $\begingroup$ HINT: Do you know that $|a_n| + |b| < C$ for $C$ independent of $n$? Are there any useful theorems that you know? $\endgroup$ – amarney Mar 22 '17 at 19:52
  • $\begingroup$ Will I also have to show cases for $b \neq 0$ and $b = 0$?. And to above, I've never heard that before. $\endgroup$ – Tinler Mar 22 '17 at 19:54
2
$\begingroup$

You can use the fact that there exists a real constant $M := \max\{|a+\epsilon|, |a-\epsilon|, |a_n|\ \mathrm{for}\ n < N\}$. Now $|a_n||b_n−b|+|b||a_n−a| ≤ M|b_n−b|+|b||a_n−a|$. You can do the rest right?

Also, at the end of the proof, you want it to say $< \epsilon$. If you start off by saying "for all $\epsilon$ there exists ... such that $|a_n-a|<\frac{\epsilon}{2|b|}$", the last line will simplify to $M|b_n−b|+|b||a_n−a| < M|b_n−b| + \frac{\epsilon}{2}$. In the same way you can make the first term simplify to $\frac{\epsilon}{2}$ and end up with a clean proof. But to do this you need to know that |b| is non-zero, so this will require a separate case.

$\endgroup$
  • $\begingroup$ Is writting $M:= sup(|a_n|)$ the same thing? $\endgroup$ – Tinler Mar 22 '17 at 19:52
  • $\begingroup$ It's not as "obvious" that the supremum exists, because you need to know it's bounded first. What I wrote is stating that it's bounded and constructing one. $\endgroup$ – Harambe Mar 22 '17 at 19:55
  • $\begingroup$ Oh okay thank you. What about when its $b = 0 $? Will I have to seperate the case? $\endgroup$ – Tinler Mar 22 '17 at 19:56
  • $\begingroup$ Oh sorry I forgot to reply to that part. If |b| is zero you don't have to do anything, you proof still works. $\endgroup$ – Harambe Mar 22 '17 at 19:57
  • $\begingroup$ Yes sorry I forgot about the /2 . $\endgroup$ – Tinler Mar 22 '17 at 20:02
1
$\begingroup$

Here is another way let $$\lim _{ n\rightarrow \infty }{ { a }_{ n }=a } ,\lim _{ n\rightarrow \infty }{ { b }_{ n }=b } $$then $${ a }_{ n }=a+{ \alpha }_{ n },{ b }_{ n }=b+{ \beta }_{ n }\quad ,\quad n=1,2,...$$ where $\lim _{ n\rightarrow \infty }{ { \alpha }_{ n }=\lim _{ n\rightarrow \infty }{ { \beta }_{ n }=0 } } $ $${ a }_{ n }{ b }_{ n }=\left( a+{ \alpha }_{ n } \right) \left( b+{ \beta }_{ n } \right) =ab+\left( { \alpha }_{ n }b+{ \beta }_{ n }a+{ \alpha }_{ n }{ \beta }_{ n } \right) \\ \lim _{ n\rightarrow \infty }{ \left( { \alpha }_{ n }b+{ \beta }_{ n }a+{ \alpha }_{ n }{ \beta }_{ n } \right) =0 } $$

so $$\lim _{ n\rightarrow \infty }{ { a }_{ n }{ b }_{ n }=ab=\lim _{ n\rightarrow \infty }{ { a }_{ n }\lim _{ n\rightarrow \infty }{ { b }_{ n } } } } $$

$\endgroup$
0
$\begingroup$

Here's a solution which mostly consists of proving several special cases, which I saw in my first analysis textbook (don't remember the title anymore):

  1. Special case where $a = b = 0$.
  2. Special case where $a_n = A$ is a constant sequence and $b = 0$.
  3. (Possibly already done) Prove if $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a_n + b_n \rightarrow a + b$ as $n \to \infty$.
  4. (Possibly already done) Prove if $a_n = A$ is a constant sequence then $a_n \to A$ as $n \to \infty$.

Then, to put all these together to prove the general case: use \begin{equation} a_n b_n = (a_n - a) (b_n - b) + a (b_n - b) + b (a_n - a) + ab. \end{equation}

$\endgroup$

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.