For reference, here is the full Theorem 3.3 (d) from Rudin.

Suppose $\{s_n\}$, $\{t_n\}$ are complex sequences, and $\lim\limits_{n \to \infty} s_n = s$, $\lim\limits_{n \to \infty} t_n = t$. Then

(a) $\lim\limits_{n \to \infty} \left(s_n + t_n\right) = s + t$;

(b) $\lim\limits_{n \to \infty} cs_n = cs$, $\lim\limits_{n \to \infty} \left(c + s_n\right) = c + s$, for any number $c$;

(c) $\lim\limits_{n \to \infty} s_n t_n = st$;

(d) $\lim\limits_{n \to \infty} \frac{1}{s_n} = \frac{1}{s}$, provided $s_n \neq 0$ ($n = 1, 2, 3, \ldots$), and $s \neq 0$.

There's one key step in the proof of (d) that I don't understand. First, Rudin uses convergence of $s_n$ to find an $m \in \mathbb{N}$ so that for all $n \geq m$, we have $|s_n - s| < \frac{1}{2} |s|$. He then asserts that for $n \geq m$, we have $$ |s_n| > \frac{1}{2} |s|. $$ This is a very important step, but I cannot follow it. I've tried contradiction and the triangle inequality, but I can't get the inequality signs to line up. For example, I tried (for $n \geq m$), $$ |s_n| = |(s_n - s) + s| \leq |s_n - s| + |s| < \frac{1}{2} |s| + |s| = \frac{3}{2} |s|. $$ It seems as though I've bounded $|s_n|$ "in the opposite direction." If I knew $s$ were positive, expanding the absolute values might work, but we only know it's non-zero.

The rest of the proof looks pretty straightforward to me, with one slight doubt. He asserts the existence of an $N$ (I don't know why he requires $N > m$ when he could just take the maximum of $N$ and $m$; does this make a difference?) so that $n \geq N$ implies $|s_n - s| < \frac{1}{2} |s|^2 \epsilon$. As $|s_n| > \frac{1}{2} |s|$, we have $\frac{1}{|s_n|} < \frac{2}{|s|}$. We then have: \begin{align*} \left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert & = \left \lvert \frac{s - s_n}{s_n \cdot s} \right \rvert \\ & = \frac{|s - s_n|}{|s||s_n|} \\ & < \frac{2|s - s_n|}{|s|^2} \\ & < \frac{2}{|s|^2} \cdot \frac{1}{2} |s|^2 \epsilon \\ & = \epsilon \end{align*}

I would appreciate some feedback on the above and some help with those two questions (how Rudin deduces $|s_n| > \frac{1}{2} |s|$ and why he takes $N > m$ instead of $\max(N,m)$.)

  • 2
    $\begingroup$ Does this answer your question? How to prove that if $\lim(a_n)=a$ then $\lim(\frac{1}{a_n})=\frac{1}{a}$? $\endgroup$
    – Mittens
    Jun 4, 2021 at 17:49
  • $\begingroup$ Try using $|a-b|\geq||a|-|b||$ $\endgroup$
    – saulspatz
    Jun 4, 2021 at 17:51
  • $\begingroup$ @saulspatz I didn't think to use that inequality. That would give $|s_n - s| \geq ||s_n| - |s||$, but I don't know where to go from there. I'm also not sure how he converts this into a strict inequality. $\endgroup$
    – JeremyS
    Jun 4, 2021 at 17:53
  • $\begingroup$ Now we have $\frac12|s|>||s_n|-|s||$ and we're only dealing with positive numbers, so it should be clear. If not, just assume the contrary. $\endgroup$
    – saulspatz
    Jun 4, 2021 at 17:57
  • 1
    $\begingroup$ I think it's worth mentioning that this has nothing to do with the signs of any of the $s_n$ or $s$, it is entirely a question about distances. The same thing holds in any normed vector space, and there, there is no concept of sign. I would argue that the right way to think about this is that once $|s_n-s|<|s|/2$, this means that $s_n$ is inside the ball of radius $|s|/2$ around $s$. But $s$ is $|s|$ away from the origin, so this means that $|s-s_n|$ is more than $|s|/2$ away from the origin. Thinking about it in terms of signs muddles the simpler and more general picture to have in mind. $\endgroup$
    – Alex Ortiz
    Jun 4, 2021 at 21:01

2 Answers 2


=== second (more obvious answer) ====

Or we could talk it out:

$|a - b| < k$ means $a$ and $b$ are within $k$ of each other. Which means $a-k < b < a+ k$.

So $s -\frac 12|s| < s_n < s+\frac 12|s|$ and so:

$\begin{cases} \frac 32 s\\0< \frac 12 s\end{cases} < s_n < \begin{cases}\frac 12s< 0&\text{if }s<0\text{ so }|s|=-s\\\frac 32s &\text{if }s> 0\text{ so }|s|=s\end{cases}$.

(In hindsight, I think Rudin was thinking this argument was obvious.... And it is obvious... in hindsight.)

(Let's face it... Rudin was just smarter than us.....)

==== first answer====

When in doubt do cases or broad strokes or.... whatever

$|s_n - s| < \frac 12 |s|$

$-\frac 12|s| < s_n -s < \frac 12 |s|$.

If $s$ is positive then $s = |s|$ and

$-\frac 12|s| < s_n - |s| < \frac 12|s|$ and

$0 < \frac 12|s| < s_n < \frac 32|s|$ and $s_n > 0$ and $|s_n| = s_n >\frac 12|s|$.

If on the on the other hand $s < 0$ then $|s| =- s$ and so

$-\frac 12|s| < s_n + |s| < \frac 12|s|$ and so

$-\frac 32|s| < s_n < -\frac 12|s| < 0$.

So $s_n < 0$ and $|s_n| = -s_n$ and $\frac 12|s| < -s_n = |s_n|$.

Either way $\frac 12|s| < |s_n| < \frac 32|s|$ and $s_n$ and $s$ are the same sign.

... There is almost certainly a slicker more direct way with triangle inequality on substituting $t = s_n -s$ so $t+s = s_n$ or something like that but nothing wrong with hitting something with a hammer until it falls apart....

We can do a clunkier way with more hammer strokes but easier calculations by considering the four cases $s_n,s$ positive or negative (as $s_n \to s$ and $s\ne 0$ we can assume $n$ is large enough to assure $s_n$ is the same sign as $s$) and whether $s_n$ is less or greater or equal to $s$...


Case 1: $0 < s < s_n$ so $s_n > s > \frac 12 s > 0$ and $0 < \frac 12|s| < |s_n|$.

Case 2: $0 < s_n \le s$ so $|s_n-s| < \frac 12|s| \implies s-s_n < \frac 12 s\implies \frac 12s < s_n \implies \frac 12|s| < |s_n|$

Case 3: $s_n < s < 0$ so $s_n < s < \frac 12 s <0$ so $0 < -\frac 12s < -s_n$ so $0< \frac 12|s| < |s_n|$

and Case 4: $s\le s_n < s < 0$ so $|s_n - s| < \frac 12s \implies 0 \le s_n - s < -\frac 12 s \implies s_n < \frac 12 s < 0\implies 0 < \frac 12|s| < |s_n|$.

... but that was pretty tedious and didn't make us look like slick rock stars.


By the triangle inequality, $|s|\le |s_n| + |s-s_n|$, or equivalently, $|s-s_n|\ge |s|-|s_n|$. For $n\ge m$ then, $|s-s_n|> \frac{|s|}{2}$.

As to your second question, there is no special reason to take $N \ge m$ or $N> m$, or to let $N'$ be such that $|s_n-s|<\frac12|s|^2\epsilon$ for $n\ge N'$, and then take $N=\max(N',m)$. The goal is simply to ensure there is some constant (which in this case is $N$ because he takes $N > m$) so that for $n$ bigger than that constant, we can ensure both inequalities we want. Taking $N > m$ is a particularly succinct way to do just that.


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