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 $0$.


Is this correct?


3 Answers 3


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*}

  • $\begingroup$ Thanks for your answer. This was copy paste error. I edited it. Is it correct now? $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 12:03
  • $\begingroup$ @Math_QED It's better, but not correct. You used implicitly (I think) that $|s_n|<1+|L_1|+|L_2|$ if $n$ is large enough; you should be explicit. Besides, aftar the last $<$ sign there is a factor $2$ missing (you are adding a thing with itself), and therefore in the end what you actually get is $2\epsilon$, not $\epsilon$. $\endgroup$ Commented Aug 10, 2017 at 12:10
  • $\begingroup$ I used $|s_n| < 1 + |L_1|$ for $n$ large enough. I wrote this down now in the proof. Hope it's correct now. Don't know what factor $2$ I'm missing though. $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 12:20
  • 1
    $\begingroup$ Thanks for all your help :). I think your proof is quite complicated by introducing minimums, but on the other hand you don't asymmetries. I prefer my proofs but that's just a matter of taste. $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 14:56
  • 1
    $\begingroup$ @puka Yes. Why not? $\endgroup$ Commented Dec 1, 2020 at 13:51

$\lim s_nt_n = \lim (s_nt_n-s_nL_2+s_nL_2) = \lim s_n(t_n-L_2)+\lim s_nL_2=(\lim s_n(t_n-L_2))+L_1L_2$.

The result comes from observing that $\lim (t_n-L_2)=0$ and that as $s_n$ is convergent it is bounded, so $\lim s_n(t_n-L_2)=0$.



there is an error in line 37.

the easiest way is to use the fact that a convergent sequence is bounded.

$$|s_n|\le S $$

and $$|s_nt_n-L_1L_2|\le $$ $$S|t_n-L_2|+|L_2||s_n-L_1|\le$$ $$S\frac {\epsilon}{2S}+|L_2|\frac {\epsilon}{2|L_2|} $$

if $$SL_2\ne 0$$

If $L_2=0$ and $S\ne 0$

$$|s_nt_n|\le S\frac {\epsilon}{S} $$ for enough large $n $.

  • $\begingroup$ This does not answer the OP's question, which is about verifying their proof. $\endgroup$ Commented Aug 10, 2017 at 11:51
  • $\begingroup$ This is actually a useful answer. $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 12:06
  • $\begingroup$ So, as an alternative answer: $s_n$ converges, so $|s_n|$ converges, so $(|s_n|)$ (the sequence of moduli of the complex numbers) is bounded, meaning there is an element $S \in \mathbb{R}$ such that $|s_n| \leq S$ for all $n$ and I can use this in the proof? $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 12:13
  • $\begingroup$ This proof does not work if either $S = 0$ or $L_2 = 0$. I followed your suggestion and added my own proof. Could you maybe look at it? $\endgroup$
    – user370967
    Commented Aug 10, 2017 at 12:35

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