For this question, I'm not sure how to do the proof. Here is what I have so far. Can anyone please help me out?
Let a, b ∈ R. Let $\{a_n\}$ and $\{b_n\}$ be sequences. Suppose that $\{a_n\}$ converges to a and that $\{b_n\}$ converges to b. Moreover suppose that {bn} is bounded above by some c ∈ R +. Provide a complete and accurate $\epsilon$ − N proof that $\{a_nb_n\}$
$\{a_n\}$ $\to$ a
$\{b_n\}$ $\to$ b
$b_n \le c$
Let $\epsilon > 0$ be arbitrary, st if $n>N1$, $|a_n - a|<\epsilon$
Let $\epsilon > 0$ be arbitrary, st if $n>N2$, $|b_n-b|<\epsilon$
N = max(N1,N2)
$|a_nb_n - ab|$
$ =|a_nb_n - a_nb + a_nb - ab|$
$= |a_n(b_n-b) + b(a_n-a)|$
$\le |a_n(b_n-b)| + |b(a_n-a)|$
$ = |a_n| |b_n-b| + b|a_n-a|$
$ < (a+\epsilon)\epsilon + b\epsilon$
$ = (a+b)\epsilon + \epsilon^2 < \epsilon$
By the triangle inequality,
$$ |a_nb_n-ab| = |a_nb_n - a_nb +a_nb - ab| \leq |a_nb_n-a_nb| + |a_nb - ab| $$
Then
$$ |a_nb_n-a_nb| + |a_nb - ab| = |a_n||b_n-b|+|b||a_n-a| $$
Now $a_n$ is bounded since it's convergent, so suppose its bounded by $M$. Thus $|a_n|\leq M$.
Now we can make $|a_n-a|$ and $|b_n-b|$ arbitrarily small. So we make $|a_n-a|<\frac{\varepsilon}{2|b|}$ and $|b_n-b|<\frac{\varepsilon}{2M}$. Then
$$ |a_nb_n-ab| \leq |a_n||b_n-b|+|b||a_n-a| < M\cdot\frac{\varepsilon}{2M} + |b|\cdot\frac{\varepsilon}{2|b|} = \varepsilon $$ which is the definition of convergence.