I'm struggling with following my professor's proof of this theorem from Ross's real-analysis textbook. I've reworked parts of it in ways that make sense to me, and it'll be a bit informal at times (especially in the choice of $N$), but I'd greatly appreciate if someone could review this proof.
Theorem: If $(s_n)$ converges to $s$, if $s_n \neq 0$ for all $n$, and if $s \neq 0$, then $\left(1/s_n\right)$ converges to $\frac{1}{s}$.
Proof: We need to show that \begin{align*} \forall \epsilon > 0, \exists N, \forall n > N, \left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert < \epsilon \end{align*} Informally, we can rewrite $\left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert$ as: \begin{align*} \left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert = \left \lvert \frac{s - s_n}{s \cdot s_n} \right \rvert \end{align*} We know that we can choose some value of $N$, which we might call $N_1$, such that $\left \lvert s - s_n \right \rvert < \frac{s}{2}$. This implies that $- \frac{s}{2} < s - s_n < \frac{s}{2}$, which implies that $s_n > \frac{s}{2}$. Thus, we have \begin{align*} \left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert = \left \lvert \frac{s - s_n}{s \cdot s_n} \right \rvert < \left \lvert \frac{s - s_n}{s \cdot \frac{s}{2}} \right \rvert = \left \lvert \frac{s - s_n}{\frac{s^2}{2}} \right \rvert = \frac{\left \lvert s - s_n \right \rvert}{\left \rvert \frac{s^2}{2} \right \rvert} \end{align*} This is then the part I'm confused on. We of course want to write this inequality chain with an $\epsilon$ in some way. We can find our goal of showing this difference is less than $\epsilon$ this by finding some way to cancel out the $\frac{s^2}{2}$. I know that we can choose some $N$, say $N_2$, such that $\left \lvert s_n - s \right \rvert < \frac{\epsilon s^2}{2}$, which is equivalent to saying that $\left \lvert s - s_n \right \rvert < \frac{\epsilon s^2}{2}$. But we've already chosen some $N$ so as to make this difference less than $\frac{s}{2}$. We can't say for certain whether $\frac{\epsilon s^2}{2}$ is greater or less than $\frac{s}{2}$. (They could also be equal if $s = \epsilon = 1$), so we can't say for certain that we're choosing $N_2$ to be strictly larger than $N_1$.
The solution that I believe works is to say this: there is some $N$, $N_1$, that makes the difference less than $\frac{s}{2}$ and some $N$, $N_2$, that makes the difference less than $\frac{\epsilon s^2}{2}$. Depending on the relationship between $\frac{s}{2}$ and $\frac{\epsilon s^2}{2}$, it could be that either $N_1 > N_2$, $N_1 < N_2$, or $N_1 = N_2$. So take $N = \max(N_1, N_2)$. This allows us to ensure both that $\left \lvert s - s_n \right \rvert < \frac{s}{2}$ and $\left \lvert s - s_n \right \rvert < \frac{\epsilon s^2}{2}$. The proof, from here, is reasonably straightforward:
For some arbitrary $\epsilon > 0$, take $N = \max(N_1, N_2)$. Then: \begin{align*} \left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert & = \left \lvert \frac{s - s_n}{s \cdot s_n} \right \rvert \\ & < \left \lvert \frac{s - s_n}{s \cdot \frac{s}{2}} \right \rvert \\ & < \left \lvert \frac{\frac{\epsilon s^2}{2}}{\frac{s^2}{2}} \right \rvert \\ & < \left \lvert \epsilon \right \rvert \\ & = \epsilon \end{align*} So, how does this look? Particularly, is my logic in choosing $N$ to be the maximum of $N_1$ and $N_2$ sound?
Thanks.
(1). The sentence "We know that .... " should say "We know there exists $N_1\in \Bbb N$ such that $|s-s_n|<s/2$ for all $n>N_1.$"
(2). Then for all $n>N_1$ we have $|1/s-1/s_n|<|s_n-s|/(|s|^2/2).$ Now for any $\epsilon^*>0$ there exists $N_2\geq N_1$ (with $N_2\in \Bbb N$) such that $|s_n-s|<\epsilon^*$ for all $n>N_2.$
Therefore, for any $\epsilon^*>0$ there exists $N_2\in \Bbb N$ such that $|1/s-1/s_n|<|s_n-s|/(|s|^2/2)<\epsilon^*/(|s|^2/2).$
(3). Now if we are given $\epsilon>0,$ we may take some (any) $\epsilon^*$ such that $0<\epsilon^*<\epsilon |s|^2/2,$ so that $\epsilon^*/(|s|^2/2)<\epsilon.$ And conclude that for any $\epsilon>0$ there exists $N_2\in \Bbb N$ such that $|1/s-1/s_n|<\epsilon^*/(|s|^2/2<\epsilon$ for all $n>N_2.$