# The quotient law for converging sequences

So I've gone through the proofs for the laws of converging sequences and I understand all of them, except the quotient law:

Let $$(a_n)$$ and $$(b_n)$$ be two sequences converging to the numbers a and b respectively. Then $$\displaylines{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}=\frac{a}{b}}$$ where b≠0.

Now I understand that in order to finish the proof, we need to prove that since $$(b_n)$$ converges to $$b$$, then $$\frac{1}{b_n}$$ converges to $$\frac{1}{b}$$.

The proof in the textbook "Understanding analysis" by Stephen Abbott is the following:

I'm totally stumped. I cannot make sense of this at all. What does the worst-case estimate mean? How come we are interested in an inequality of the form mentioned above and how does this lead to a bound on the size of $$\frac{1}{|b||b_n|}$$. Why do the terms in the sequence have to be closer to b than 0. Can somebody explain this proof in simple words please?

Now I know the same question has already been asked here, but I don't really understand the answers there either.

• Often analysis focusses on bounding quantities from above; however, when dealing with decreasing functions like $1/x$, to bound them below some $\epsilon$ requires bounding their denominators above some $\delta$, because when the reciprocal is taken that’ll result in a less-than. In short, the reciprocal makes our greater-than turn into a less-than, which is our goal. If we tried a less-than at the beginning, the reciprocal would make it greater-than $\epsilon$, which is useless to us Nov 27, 2021 at 16:38

"Worst-case" just means to find an upper bound.

Do you understand how he gets the estimate that for all $$n\geq N_1$$, we have $$|b_n|>\frac{|b|}{2}$$? If yes, then you just need to remember that when taking reciprocals, the direction of inequalities "flip". Therefore, $$\frac{1}{|b_n|}<\frac{2}{|b|}$$. We can now multiply both sides of this inequality by the positive number $$\frac{1}{|b|}$$. Multiplying by positive numbers doesn't change the inequality sign, so $$\frac{1}{|b|}\cdot \frac{1}{|b_n|}<\frac{1}{|b|}\cdot \frac{2}{|b|}$$, or rewriting this in the usual way, $$\frac{1}{|b_n||b|}<\frac{2}{|b|^2}$$.

The idea of the proof is that our assumption tells us $$b_n\to b$$, meaning we only have good control of terms involving $$|b_n-b|$$ (this is essentially what the definition of convergence is). Now, you should ALWAYS keep in mind the following heuristic principle small times manageable = small.

So, now when we want to prove convergence for the reciprocal, we have to somehow get terms like $$|b_n-b|$$ to appear, so that we can invoke our hypothesis. In this case, we can do so simply with some algebra. For all $$n\geq N_1$$, we have $$|b_n|>\frac{|b|}{2}>0$$, so $$b_n\neq 0$$, so it actually makes sense to take its reciprocal. Now, simple algebra shows us \begin{align} \left|\frac{1}{b_n}-\frac{1}{b}\right|&=|b_n-b|\cdot \frac{1}{|b_n||b|} \end{align} The first term $$|b_n-b|$$ is what we have very good control over; I can make it as small as I want by making $$n$$ large enough. The second term on the other hand is not necessarily a small quantity. But, that doesn't matter because I have the manageable estimate $$\frac{1}{|b_n||b|}<\frac{2}{|b|^2}$$. Now is where we invoke the principle that small times manageable is still small.

This is what happens in the proof by saying that given $$\epsilon>0$$, there is some $$N_2$$ such that if $$n\geq N_2$$ then $$|b_n-b|<\frac{\epsilon}{2/|b|^2}=\frac{\epsilon |b|^2}{2}$$. (This is saying $$|b_n-b|$$ is small provided $$n\geq N_2$$). Therefore, by combining the two estimates, we get that for all $$n\geq \max(N_1,N_2)$$, \begin{align} \left|\frac{1}{b_n}-\frac{1}{b}\right|&=|b_n-b|\cdot \frac{1}{|b_n||b|}<\frac{\epsilon |b|^2}{2}\cdot \frac{2}{|b|^2}=\epsilon. \end{align}

You mentioned that you understood all the laws for converging sequences. So, I suggest you look over the proof for products, and try to identify where the above principle of "small times manageable= small" was used. Perhaps this will help you understand better the proof idea for quotients.

Finally, I should mention that there's actually no need to get a "perfect $$\epsilon$$" at the end. In other words, the following is also an equivalent condition for convergence (each $$\alpha_n,\alpha$$ is assumed to be a real/complex number):

A sequence $$(\alpha_n)_{n=1}^{\infty}$$ converges to $$\alpha$$ if and only if there is a $$K\geq 0$$ such that for all $$\epsilon>0$$, there is some $$N\in\Bbb{N}$$, such that for all $$n\in\Bbb{N}$$, if $$n\geq N$$, then $$|\alpha_n-\alpha|\leq K\epsilon$$.

I leave it to you to figure out the equivalence.