# 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:

1. What am I missing in the last step? If $$x>a$$ and $$x\geq b$$ why should it always be $$b\leq\ a$$?
2. 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.

• Are you sure it isn't $(x>a\land x\ge b)\implies b\color{red}{\ge }a$? Oct 3 '20 at 17:02
• That's indeed what buzzes me..
– erma
Oct 3 '20 at 17:14
• @erma: To get a centred displayed expression, enclose it in double dollar signs: $$expression$$. Click on edit below the question to see what I did. Oct 3 '20 at 18:15

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.

• answer 1) I followed your proof and I am fine with the first part, but if we then replace $a_n<\beta$ with $a_n\leq\beta$ and come to the conclusion that $a_{n}\leq\beta$ for $n>N$ with $\beta > \alpha$ aren't we contradicting the definition of $\alpha$ by including the $\alpha =\beta$ case? I think at this point we should change also $\beta > \alpha$ in $\beta\geq\alpha$.
– erma
Oct 4 '20 at 9:11
• answer 2) I can't choose $\beta$ such that $\gamma>\beta>\alpha$ because $\beta\geq\gamma$ must hold.
– erma
Oct 4 '20 at 9:14
• Answer 1: The claim does not hold if you allow $\alpha=\beta$. Take, for example, $a_n=(-1)^n \frac{1}{n}$. This converges to $0$ so $\alpha=0$. But it is not true that there exists $N$ such that $a_n\leq \alpha$ for all $n\geq N$. The claim, in full, generality, requires $\alpha<\beta$. Oct 4 '20 at 13:16
• Answer 2: That is precisely the contradiction. If $\gamma >\alpha$ was assumed (proof by contradiction). Then it is true that there exists a number between $\gamma$ and $\alpha$. Call it $\beta$. But then this contradicts the assumption that $\beta \geq \gamma$ for all $\beta>\alpha$. Oct 4 '20 at 13:17
• Answer 1) I agree that we want $\alpha < \beta$. I was pointing out that by replacing $a_n<\beta$ with $a_n\leq\beta$ (and knowing that $a_n\leq\alpha$) we are not excluding $\alpha=\beta$ which is not what we want.
– erma
Oct 4 '20 at 21:00

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.

• In the first part, you stated $\beta= \frac{\alpha + \gamma}{2}<\gamma$ violating what is found in the proof (i.e. $\beta\leq\gamma$) and then you used $\beta\leq\gamma$ to find that it cannot be $\gamma>\alpha$. How is it fair?
– erma
Oct 4 '20 at 21:45
• @erma In the proof we found that for each $\beta > \alpha$ we have $\gamma \le \beta$ (not $\beta \le \gamma$ ). Then we prove by contradiction that $\gamma \le \alpha$: Assuming that $\gamma > \alpha$, then certainly $\beta := \alpha + \frac{\gamma - \alpha}{2} > \alpha$. This implies $\gamma \le \beta = \alpha + \frac{\gamma - \alpha}{2}$ which is not true. Oct 4 '20 at 23:19
• Yeah, I miss wrote, I shouldn't reply when tired! So first you suppose $\gamma>\alpha$ and find that this can imply $\gamma>\beta$ which is against the proof so it had to be $\gamma\leq\alpha$.
– erma
Oct 5 '20 at 16:16
• @erma Correct. If $\gamma > \alpha$, we can alternatively pick any $\beta$ such that $\alpha < \beta < \gamma$. Then by our proof $\gamma \le \beta$ which is a contradiction. Oct 5 '20 at 17:12
• Great. Thanks Paul, much appreciated.
– erma
Oct 6 '20 at 19:11