I want to prove that if:

$$\lim_{n \to \infty}s_n = L_1, \lim_{n \to \infty}t_n = L_2$$

then $$\lim_{n \to \infty}(s_n t_n) = L_1L_2$$

Wrong (?) proof:

Fix $\epsilon >0$. By definition, there are integers $N_1,N_2$ such that:

$$n>N_1 \implies |s_n-L_1|< \frac{\epsilon}{|s_n|+|L_2|}$$ $$n>N_2 \implies |t_n-L_2|< \frac{\epsilon}{|s_n|+|L_2|}$$

Hence, for $n > \max\{N_1,N_2\}$, we have:

$$|s_nt_n - L_1L_2| = |s_n(t_n - L_2) + s_nL_2 - L_1L_2|$$ $$\leq |s_n||t_n - L_2| + |L_2||s_n - L_1|$$ $$< \frac{\epsilon}{|s_n|+|L_2|} (|s_n| + |L_2|) = \epsilon$$

I was taught that the $\epsilon$ can't depend on $n$, but I can't see why. What goes wrong?

EDIT: I know how to fix the proof, I made a post on this one: Limit of product of sequences is the product of the limits of the sequences

  • $\begingroup$ The proof can be corrected by noticing that the sequence $(s_n)_{n=1}^\infty$ is bounded, since it is convergent. Take $M > 0$ such that $|s_n|\leq M$, $\forall n\in\mathbb{N}$. Now use the definition of limit for $\frac{\varepsilon}{M + |L_2|}$, as it does not depend on $n$. $\endgroup$ Sep 10, 2017 at 11:36
  • $\begingroup$ I don't exactly know whether it is wrong. Everyone says how I can correct it, but I know this (see link in post). I was just curious why this doesn't work. $\endgroup$
    – user370967
    Sep 10, 2017 at 11:41
  • $\begingroup$ Oh, I had to write out to understand. Did any of the answers below help you? If you still need it, I'll try to help. $\endgroup$
    – Git Gud
    Sep 10, 2017 at 11:48
  • $\begingroup$ I just can't see how the limit definition $\forall \epsilon >0: \exists N \in \mathbb{N}: (n>N \implies |s_n - L| < \epsilon)$ forbids the dependence of $n$ in the $\epsilon$. ajotatxe showed with an example why this isn't possible, though $\endgroup$
    – user370967
    Sep 10, 2017 at 11:51
  • 1
    $\begingroup$ @Math_QED The limit defintion should rather be $\forall\epsilon>0\colon\exists N\in\Bbb N\colon \forall n\in\Bbb N\colon (n>N\implies |s_n-L|<\epsilon)$. Since the $\forall n$ is "inside" the $\forall \epsilon$, $\epsilon$ must not depend on $n$. $\endgroup$ Sep 10, 2017 at 14:49

4 Answers 4


I always wonder why people insist in cutting epsilons in pieces.

Just remember that eventually proving $\forall n>N, |u_n-\ell|<K\varepsilon$ is sufficient as long as $K$ is constant.

Trying to cut epsilons is maybe aesthetically nice but, I think it hurts the understanding at basic level.

Instead with the straightforward proof you get to : $\begin{cases} \forall n>N_1, |s_n-L_1|<\varepsilon\\ \forall n>N_2, |t_n-L_2|<\varepsilon\\ \end{cases}$

So for $N>\max(N_1,N_2)$

we have $|s_nt_n-L_1L_2|\le|s_n||t_n-L_2|+|L_2||s_n-L_1|\le\left(|s_n|+|L_2|\right)\varepsilon$

Now you see that you do not have a constant before $\varepsilon$, and get to think about why $(s_n)_n$ should be bounded.

And indeed, any convergent sequence is bounded, thus $|s_n|<M$ independently of $n$.

You arrive to $|s_nt_n-L_1L_2|<\underbrace{(M+|L_2|)}_{\text{a constant }K}\varepsilon$

And you should be happy with that, it is not mandatory to get to a bare $\varepsilon$ in the end, $0.0003\,\varepsilon,\ 210734\,\varepsilon,\ 10^{513}\,\varepsilon$ or $K\varepsilon$ are all the same.

  • 1
    $\begingroup$ I also figured out that $<K\epsilon$ is good enough: math.stackexchange.com/questions/2390053/… ; Thank you for the answer though, it provides a lot of insight. $\endgroup$
    – user370967
    Sep 10, 2017 at 12:11
  • $\begingroup$ I believe it is a tradition perhaps started by some formalist who paid attention to form rather than essence. $\endgroup$
    – Paramanand Singh
    Sep 11, 2017 at 11:37


Take the sequence $a_n=1/n$. It converges to zero, but the statement "There exists some $N\in \Bbb N$ such that $n>N$ implies $|a_n|<\epsilon/n^2$" is false. So, indeed, $\epsilon$ can't depend on $n$.

Fixing the proof:

To fix your proof, take an upper bound $M$ of $|s_n|$, which must exist because $s_n$ converges, and write $M$ instead of $|s_n|$ in those denominators.


Your proof is almost correct. To avoid $n$- dependence and zero denominators just replace the bound $\dfrac{\epsilon}{|s_n|+|L_2|}$ with $\dfrac{\epsilon}{M+|L_2|}$ where $M$ is a real positive number such that $|s_n|<M$ for all $n$. Since $(s_n)_n$ is convergent it is bounded and therefore the number $M$ exists.


The limit of a sequence is a (topological) tool (very useful, indeed) that's needed to say whether a sequence is definitely in any neighborhood of the limit (also known as convergence point). So to check this for a given sequence you must fix a neighborhood (that is, $\varepsilon$) and then verify that for sufficiently large $n$ all the remaining terms of the sequence belongs to the fixed neighborhood. After that you must repeat the same process for all the neighborhood of the limit (this account for the "$\forall \varepsilon$" part of the definition).

  • $\begingroup$ There is no need to repeat the process, but just ensure that the process works for any positive $\epsilon$ and does not in any way depend on $\epsilon$ having some specific value. $\endgroup$
    – Paramanand Singh
    Sep 11, 2017 at 11:41
  • $\begingroup$ @ParamanandSingh thanks for your comment. "ensure that the process works for any positive $\varepsilon$" is what I mean by "repeat the same process for all the neighborhood of the limit". So we are saying the same thing. As with all processes of mathematical analysis, taking the limit is an infinite process. $\endgroup$
    – trying
    Sep 11, 2017 at 13:30

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