Why is this proof incorrect? (limit product is product of the limits)

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

Example:

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.

Answer 4

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).