I want to prove that if:
$$\lim_{n \to \infty}s_n = L_1, \lim_{n \to \infty}t_n = L_2$$
then $$\lim_{n \to \infty}(s_n t_n) = L_1L_2$$
where $s_n, t_n$ are complex sequences (I'm working through Rudin's baby rudin, and I did this proof slightly different than him, so I would like it if someone took the time to look through my proof and verify whether it's correct)
My attempt (scratch proof: finding the appropriate epsilons etc):
By definition of limit, there are positive integers $N_1,N_2$ such that:
$$n >N_1 \Rightarrow |s_n – L_1| <\epsilon'$$
$$n >N_2 \Rightarrow |t_n – L_2| <\epsilon''$$
where we can freely choose $\epsilon', \epsilon'' >0$
Let $\epsilon > 0$
Now, if $n > \max\{N_1,N_2\}$, then:
$$|s_nt_n – L_1L_2| = |s_n(t_n – L_2) + s_nL_2 – L_1L_2|$$
$$\leq |s_n||t_n – L_2| + |L_2||s_n – L_1|$$
$$\leq |s_n|\epsilon'' + |L_2|\epsilon'$$
We want this expression to be smaller than $\epsilon$, but we can't make our $\epsilon$ depend on $n$, therefore we have to find an upper bound for $|s_n|$
There is a positive integer $N_3$ such that:
$$n > N_3 \Rightarrow |s_n – L_1| < 1$$
$$\Rightarrow |s_n| < 1 + |L_1|$$
so for $n > \max\{N_1,N_2,N_3\}$, we have:
$$ |s_n|\epsilon'' + |L_2|\epsilon' < (1 + |L_1|)\epsilon'' + |L_2|\epsilon'$$
and by chosing $\epsilon'' = \epsilon' = \frac{\epsilon}{1+|L_1|+|L_2|}$, the expression becomes smaller than $\epsilon$
Rigorous proof
Let $\epsilon > 0$
By definition of limit, there are positive integers $N_1,N_2, N_3$ such that:
$$n >N_1 \Rightarrow |s_n – L_1| <\frac{\epsilon}{1+|L_1|+|L_2|}$$
$$n >N_2 \Rightarrow |t_n – L_2| <\frac{\epsilon}{1+|L_1|+|L_2|}$$
$$n > N_3 \Rightarrow |s_n – L_1| < 1 \Rightarrow |s_n| < 1 + |L_1|$$
and for $n > \max\{N_1,N_2,N_3\}$, we have:
$$|s_nt_n – L_1L_2| = |s_n(t_n – L_2) + s_nL_2 – L_1L_2|$$
$$\leq |s_n||t_n – L_2| + |L_2||s_n – L_1|$$
$$< (1 + |L_1|)\frac{\epsilon}{1+|L_1|+|L_2|} + |L_2|\frac{\epsilon}{1+|L_1|+|L_2|} = \frac{\epsilon}{1+|L_1|+|L_2|}(1+ |L_1| + |L_2|) = \epsilon$$
Alternative proof:
Let $\epsilon > 0$
$(s_n)$ converges, so $(|s_n|)$ converges and it is bounded, meaning that there is a positive real number $S$ such that $|s_n| \leq S$ for every positive $n$.
By definition of limit, there are positive integers $N_1,N_2$ such that:
$$n >N_1 \Rightarrow |s_n – L_1| <\frac{\epsilon}{S + |L_2|}$$
$$n >N_2 \Rightarrow |t_n – L_2| <\frac{\epsilon}{S + |L_2|}$$
Now, for $n > \max\{N_1,N_2\}$, we have:
$$|s_nt_n – L_1L_2| = |s_n(t_n – L_2) + s_nL_2 – L_1L_2|$$
$$\leq |s_n||t_n – L_2| + |L_2||s_n – L_1|$$
$$< (S+|L_2|)\frac{\epsilon}{S + |L_2|} = \epsilon $$
Note that this wouldn't work whenever $S + |L_2| = 0$. However, if $S=0$, then $s_n$ is the null-sequence and the theorem is trivial, so we can safely assume $S \neq 0$ such that the denominator never becomes $0$.
QED
Is this correct?
Best Answer
Your “Rigorous proof” is not correct. Look at the way you chose $N_1$, $N_2$, and $N_3$. Only the sequence $(s_n)_{n\in\mathbb N}$ is mentioned, not the sequence $(t_n)_{n\in\mathbb N}$. You can't get a correct proof without using both of them.
Here is how I would prove the same statement. Please keep in mind that I always try to avoid to introduce asymmetries when the hypothesis are symmetric. In this case, what is supposed about both sequences is the same and therefore I will avoid to treat them in two different ways.
For each $n\in\mathbb N$, you know that$$s_nt_n-L_1L_2=(s_n-L_1)L_2+(t_n-L_2)L_1+(s_n-L_1)(t_n-L_2)$$and therefore that$$|s_nt_n-L_1L_2|\leqslant|s_n-L_1|.|L_2|+|t_n-L_2|.|L_1|+|s_n-L_1|.|t_n-L_2|.$$Let $\varepsilon>0$ and choose $N_1\in\mathbb N$ such that$$n\geqslant N_1\Longrightarrow|s_n-L_1|<\min\left\{\frac\varepsilon{3\bigl(|L_2|+1\bigr)},\sqrt{\frac\varepsilon3}\right\}.$$Also, choose $N_2\in\mathbb N$ such that$$n\geqslant N_2\Longrightarrow|t_n-L_2|<\min\left\{\frac\varepsilon{3\bigl(|L_1|+1\bigr)},\sqrt{\frac\varepsilon3}\right\}.$$Let $N=\max\{N_1,N_2\}$. Then, if $n\geqslant N$,\begin{multline*}|s_n-L_1|.|L_2|+|t_n-L_2|.|L_1|+|s_n-L_1|.|t_n-L_2|<\\<\frac{\varepsilon|L_2|}{3\bigl(|L_2|+1\bigr)}+\frac{\varepsilon|L_1|}{3\bigl(|L_1|+1\bigr)}+\frac\varepsilon3<\varepsilon.\end{multline*}