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

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

(a) $limlimits_{n to infty} left(s_n + t_nright) = s + t$;

(b) $limlimits_{n to infty} cs_n = cs$, $limlimits_{n to infty} left(c + s_nright) = c + s$, for any number $c$;

(c) $limlimits_{n to infty} s_n t_n = st$;

(d) $limlimits_{n to infty} frac{1}{s_n} = frac{1}{s}$, 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 $|s_n – s| < frac{1}{s} |s|$. 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 $|s_n| > frac{1}{s} |s|$, 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 $max(N,m)$.)