Baby Rudin th 3.37: some thoughts at the very final step I am pretty new here...first of all, how do I center text? I couldn't center the mathematical steps, I am willing to edit this!
Now to the point, theorem 3.37 of Rudin's PMA states:

3.37 Theorem For any sequence {$c_{n}$} of positive numbers,
$$\liminf_{n \to \infty} \frac{c_{n+1}}{c_n} \leq \liminf_{n \to \infty} \sqrt[n]{c_n} \\ \limsup_{n \to \infty} \sqrt[n]{c_n} \leq \limsup_{n \to \infty} \frac{c_{n+1}}{c_n}.$$
Proof We shall prove the second inequality; the proof of the first is quite similar. Put
$$\alpha = \limsup_{n \to \infty} \frac{c_{n+1}}{c_n}.$$
If $\alpha=+\infty$, there is nothing to prove. If $\alpha$ is finite, choose $\beta>\alpha$. There is an integer $N$ such that
$$\frac{c_{n+1}}{c_n} \leq \beta$$
for $n≥N$. In particular, for any $p>0$,
$$c_{N+k+1} \leq \beta c_{N+k} \;(k=0,1,...,p-1).$$
Multiplying these inequalities, we obtain
$$c_{N+p} \leq \beta^p c_N,$$
or
$$c_n \leq c_N \beta^{-N} \cdot \beta^n \quad (n \geq N).$$
Hence
$$\sqrt[n]{c_n} \leq \sqrt[n]{c_N \beta^{-N}} \cdot \beta,$$
so that
$$\limsup_{n \to \infty} \sqrt[n]{c_n} \leq \beta, \quad \quad (18)$$
by Theorem 3.20(b). Since (18) is true for every $\beta>\alpha$, we have
$$\limsup_{n \to \infty} \sqrt[n]{c_n} \leq \alpha.$$

All the steps are clear to me except the very last one:
if $\beta>\alpha$ and $\limsup_{n \to \infty} \sqrt[n]{c_n}=\gamma \leq \beta$ why is it always $\gamma\leq\alpha$? Even if $\beta=\alpha+d$ with d infinitesimally small, if $\gamma=\beta\leq\beta$ it results in $\gamma > \alpha$ right?
To me the problem is indeed the "$\leq$" instead of "$<$" in $\gamma\leq\beta$: if it was $\gamma<\beta$ then I can always choose $\beta=\alpha+d$ so that $\gamma<\beta$ results in $\gamma\leq\alpha$ (since $\alpha<\beta$). This is the only way I could think of it intuitively even if using the "infinitesimally small" argument in such a rough way could be formally wrong.
This said I proceeded and looked back where "$\leq$" appears for the first time and it is at the second step of the proof, when it's stated that
$\frac{c_{n+1}}{c_n} \leq \beta$. Now, I guess this comes from theorem 3.17b) which says that (using the above notation)

If $\beta>\limsup_{n \to \infty} \frac{c_{n+1}}{c_n}$, there is an integer N such that $n\geq N$ implies $\frac{c_{n+1}}{c_n}<\beta$

and not $\frac{c_{n+1}}{c_n}\leq\beta$.
So my questions are:

*

*What am I missing in the last step? If $x>a$ and $x\geq b$ why should it always be $b\leq\ a$?

*Why Rudin uses "$\leq$" instead of $<$ in the second step, provided it comes from Theorem 3.17?

Thanks in advance, I hope my thoughs were clearly explained.
(let me know how to center the text!)

EDIT: I found out that user @Mikhail D had the same flow of thoughts from theorem 3.17 to 3.37. He explained it more organically than what I did as answer of the following post.
 A: Let $\alpha = \limsup_{n \to \infty} \frac{c_{n+1}}{c_n}$ and $\gamma = \limsup_{n \to \infty} \sqrt[n]{c_n}$. Rudin wants to show that
$$\gamma \leq \alpha . \tag{1}$$
You say that all the steps are clear to you except the very last one. My interpretation is that you accept (18) which means that
$$\gamma \leq \beta \text{ for } \textbf{ each } \beta > \alpha . \tag{2}$$
Now assume that $\gamma > \alpha$. Then we may choose $\beta = \alpha + \frac{\gamma - \alpha}{2}$ and obtain from (18)
$$\gamma \le \alpha + \frac{\gamma - \alpha}{2} $$
which is equivalent to $2\gamma < \gamma$, a contradiction. Therefore $(1)$ is true.
Of course you are right that if we are given three numbers $\alpha, \beta, \gamma$ such that $\beta > \alpha$ and $\gamma \le \beta$, then we cannot conclude that $\gamma \le \alpha$. But that is not the situation here: In fact we have $\gamma \le \beta$ for all $\beta > \alpha$.
Finally, you can of course start the proof with the stronger (but correct) fact

There is an integer $N$ such that $\frac{c_{n+1}}{c_n} < \beta$ for $n \ge N$.

This gives you
$$\sqrt[n]{c_n} < \sqrt[n]{c_N \beta^{-N}} \cdot \beta .$$
However, this does not imply $\limsup_{n \to \infty}\sqrt[n]{c_n} < \beta$, but only $\limsup_{n \to \infty}\sqrt[n]{c_n} \leq \beta$ as stated by Rudin.
A: First of all, from the first step, we can replace $\leq$ with $<$ if that makes the proof easier. Why?
Claim 1: Let $\{a_n\}$ be a bounded sequence with $\alpha=\limsup a_n$. Then for all $\beta>\alpha$, there exists $N$ such that $a_n<\beta$ for all $n\geq N$.
Proof: Let $\beta>\alpha$ and suppose for each $k$, there exists $n_k>k$ such that $a_{n_k}\geq \beta$. Any convergent subsequences of $\{a_{n_k}\}$ must be a convergent subsequence of $\{a_n\}$. Hence, $\alpha\geq \limsup_{k\to\infty} a_{n_k}\geq \beta$, which is a contradiction.
Now, if you run through the same proof, but replace $a_n<\beta$ with $a_n\leq \beta$, the conclusion is the same. Thus, it doesn't really matter which one you use.
As far as the claim: If $\gamma \leq \beta$ for all $\beta>\alpha$, then $\gamma \leq \alpha$.
Proof: Suppose not; that is, $\gamma >\alpha$. Choose $\beta$ such that $\gamma>\beta>\alpha$. Then $\gamma >\beta$ and $\gamma \leq \beta$, a contradiction.
