# Understanding Rudin Theorem 3.3 (d)

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 $$\max(N,m)$$.)

• Does this answer your question? How to prove that if $\lim(a_n)=a$ then $\lim(\frac{1}{a_n})=\frac{1}{a}$? Jun 4, 2021 at 17:49
• Try using $|a-b|\geq||a|-|b||$ Jun 4, 2021 at 17:51
• @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. Jun 4, 2021 at 17:53
• 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. Jun 4, 2021 at 17:57
• 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. Jun 4, 2021 at 21:01

=== 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.....)

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 $$s$$...

i.e.

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.