sequences and series - Another Simple convergence theorem proof

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 $a_nb_n \Rightarrow LM$

Solution: (My attempt)

WTS:

Let R = sup(|$a_n$|)

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

$|a_n - L| < \frac{\epsilon}{(R + |M|)}$

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

$|b_n - M| < \frac{\epsilon}{(R + |M|)}$

Choose $N = \max(N_1, N_2)$

Suppose $n > N$

$$|a_nb_n - LM| = |a_nb_n - a_nM + a_nM - LM|$$

$$= |a_n(b_n - M) + M(a_n - L)| \text{ by algebra}$$

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

$$< (R + |M|)\frac{\epsilon}{(R + |M|)}$$

$$= \epsilon$$

I have asked this before but I'm wondering if this way to write it is correct as well because this is the style my teacher does it.

Answer

Your last inequality

$$(|a_n|+|M|)\epsilon<\epsilon$$

is not correct. But if you start your proof with $$|a_n-L|<\frac{\epsilon}{R+|M|},\quad |b_n-M|<\frac{\epsilon}{R+|M|}$$

where $R:=\sup |a_n|$, then in your step previous to the end you will had

$$(|a_n|+|M|)\frac{\epsilon}{R+|M|}\le(R+|M|)\frac{\epsilon}{R+|M|}=\epsilon$$

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Note that you can define $R$ for any convergent sequence because if $(a_n)$ is bounded then $(|a_n|)$ is also bounded, and then $R$ exists.