2

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.

erma
  • 87
  • 7

2 Answers2

2

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.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125
  • 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 04 '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. – Paul Frost Oct 04 '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 05 '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. – Paul Frost Oct 05 '20 at 17:12
0

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.

ProfOak
  • 5,915
  • 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 04 '20 at 09:11
  • answer 2) I can't choose $\beta$ such that $\gamma>\beta>\alpha$ because $\beta\geq\gamma$ must hold. – erma Oct 04 '20 at 09:14
  • 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 $\alpha<\beta$. – ProfOak Oct 04 '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 $\beta>\alpha$. – ProfOak Oct 04 '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 04 '20 at 21:00
  • Answer 2) What I can't understand is the following: we arrived to the conclusion (through the proof) that $\gamma\leq\beta$ and then it appears to me that what you are suggesting is picking $\beta$ in such a way it violates $\gamma\leq\beta$ which cannot be since we proved the opposite. – erma Oct 04 '20 at 21:20
  • I will not respond to answer 1, as I feel like I will just confuse you more. My recommendation is the following. Re-read Rudin and the theorems leading up to this problem. You will find that it doesn't matter whether one uses $<$ or $\leq$. I wouldn't get too focused on this. – ProfOak Oct 04 '20 at 22:19
  • Regarding answer 2), what Rudin showed was that $\gamma \leq \beta $ for all $\beta>\alpha$. Your difficulty was understanding why this fact implies $\gamma\leq \alpha$. I proved this implication by supposing it wasn't true. Then $\gamma>\alpha$. But if this is true, pick any $\beta$ with $\gamma>\beta>\alpha$. This is a contradiction because it was just shown that $\gamma\leq \beta$ for any $\beta>\alpha$. This contradiction shows that $\gamma\leq \alpha$. – ProfOak Oct 04 '20 at 22:21
  • zugzug thanks for your patience. With a clear mind today I got your point on answer 2) and yes I need to re-study some of the previous theorems. – erma Oct 05 '20 at 16:07
  • One last thing. In your main post you say "...replace $a_n<\beta$ with $a_n\leq\beta$, the conclusion is the same...". I do agree, but I was thinking not of that inequality but $\alpha<\beta$ vs $\alpha\leq\beta$ instead. That would be a problem right? Because in the proof it appears $\alpha\geq \limsup_{k\to\infty} a_{n_k}\geq \beta\geq\alpha$ which would admit no contradiction for $\alpha=\beta$. And thus I agree on what you said below that "The claim, in full generality, requires $\alpha<\beta$." – erma Oct 05 '20 at 16:29
  • Yes, you are correct. In the claim that I proved above, it is very important that $\alpha<\beta$. See the example in the comment above where $a_n$ oscillates around $0$ and converges to $0$. Clearly, there is never a point where $a_n <0$ or $a_n \leq 0$ for all $n\geq N$. – ProfOak Oct 05 '20 at 16:33