They start by choosing $m$ s.t. $|s_n - s| \lt \frac{1}{2} |s|$ if $n > m$. From here, it looks like they use the triangular inequality $|s_n - s| + |s_n| < |s|$ to come up with this statement but I'm not sure as they introduce some variable m.

Next, given $\epsilon > 0$, there is a $N > m$ s.t $n \geq N$ (why not just state this in the first place?) implies $|s_n - s| < \frac{1}{2}|s^2|\epsilon$.

Hence for $n \geq N$ we have:

$|\frac{1}{s_n} - \frac{1}{s}| < |\frac{s_n-s}{s_ns}|<\frac{2}{|s^2|}|s_n-s|<\epsilon$.

I can follow the inequalities, but I'm not sure what techniques they used to come to this conclusion. I would appreciate it if someone could shed some light on this, thanks!

  • $\begingroup$ Not only they want $|s_n-s|$ to be bounded by $1/2 |s|^2 \epsilon$ but they also want $|s_n|$ to be bounded by $1/2|s|$ for this they get another $N_0$ which they are calling $m$. So in a nutshell they need two inequalities out of the convergence of the sequence, and they are getting two cut offs. $\endgroup$ Aug 6, 2015 at 6:00

2 Answers 2


We want to prove that $\frac{1}{s_n} \rightarrow \frac{1}{s}$. Okay, let's evaluate:

$\displaystyle \lvert\frac{1}{s_n} - \frac{1}{s}\rvert=\lvert\frac{s -s_n}{s_ns}\rvert$

We know what is upside converges to $0$... but what is down may be tricky to deal with. It could also go to $0$! But the idea is that, since $s_n \rightarrow s$, then what is down will actually go to $s^2$, so we will get $\frac{0}{s^2}=0$. Note, therefore, that it is enough to prove that $|ss_n|$ is bounded by below, at least eventually... which we will conclude from the fact that $ss_n \sim s^2$

Okay, since $s_n \rightarrow s$, then $ss_n \rightarrow s^2$, and $|ss_n| \rightarrow s^2$. Since we want to bound $ss_n$ by below, and this tends to $s^2$, which is different than $0$ by hypothesis, it is very tempting to take $s^2/2$ as a lower bound. Doing this, we have that for large enough $n$, we get $|ss_n| > \frac{s^2}{2}$, which implies that for large enough $n$, $\frac{1}{|ss_n|} < \frac{2}{s^2}$.

Well, since we can control the numerator, just adjust the $\epsilon$ to suit our needs... for large enough $n$, we have that $|s-s_n| < \frac{1}{2}s^2\epsilon$, and the $\epsilon$'s adjusts just fine at the conclusion.

Rudin is a book that most of the times wants to give a rigorous and succinct proof. As such, it is a good exercise to look for "how he got this idea?".

  • 1
    $\begingroup$ I tried working backwards as well.. but this is perfect! $\endgroup$
    – Ken
    Aug 6, 2015 at 6:48

This bit can be proved as follows.

By assumption $s_n \to s$. Using the Reverse triangle inequality (see Wikipedia, or Rudin Ex 13 Chapter 1 on page 23)

$$ \left| |s_n| - |s| \right| \le \left| s_n - s \right| $$

So there exists such M' that if $n \ge M'$ implies

\begin{align} & |s| - |s_n| < \frac{1}{2}|s| \nonumber \\ & \frac{1}{2} |s| < |s_n| \nonumber \\ & \frac{1}{2} |s|^2 < |s_n s| \nonumber \qquad \text{(A)} \end{align}

Similarly, for any number $\frac{1}{2} |s|^2 \epsilon$ we can find $M''$ such that if $n \ge M''$ then

$$ |s_n - s| < \frac{1}{2} |s|^2 \epsilon \qquad \text{(B)} $$

Now we use (A) and (B)

$$ \left| \frac{1}{s_n} - \frac{1}{s} \right| = \underbrace{ \left| \frac{s - s_n}{s_n s} \right| }_\text{"Left"} < \underbrace{ \frac{\frac{1}{2}|s|^2\epsilon}{\frac{1}{2}|s|^2} }_\text{"Right"} = \epsilon $$

In the preceding expression "Right" is (B) divided by (A), which are both positive for $\epsilon > 0$. Nominator of "Right" is greater than that of "Left". Denominator of "Right" is less than that of "Left".


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