# 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

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

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

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

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

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

• 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...) – Max Mar 22 '17 at 19:49
• 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. – Tinler Mar 22 '17 at 19:52
• HINT: Do you know that $|a_n| + |b| < C$ for $C$ independent of $n$? Are there any useful theorems that you know? – amarney Mar 22 '17 at 19:52
• Will I also have to show cases for $b \neq 0$ and $b = 0$?. And to above, I've never heard that before. – Tinler Mar 22 '17 at 19:54

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.

• Is writting $M:= sup(|a_n|)$ the same thing? – Tinler Mar 22 '17 at 19:52
• 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. – Harambe Mar 22 '17 at 19:55
• Oh okay thank you. What about when its $b = 0$? Will I have to seperate the case? – Tinler Mar 22 '17 at 19:56
• Oh sorry I forgot to reply to that part. If |b| is zero you don't have to do anything, you proof still works. – Harambe Mar 22 '17 at 19:57
• Yes sorry I forgot about the /2 . – Tinler Mar 22 '17 at 20:02

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

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 $$a_n b_n = (a_n - a) (b_n - b) + a (b_n - b) + b (a_n - a) + ab.$$