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:


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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.

  • $\begingroup$ 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 $\endgroup$
    – FShrike
    Nov 27, 2021 at 16:38

1 Answer 1


"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.


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