2
$\begingroup$

I am looking for some help with making sense of the proof in Baby Rudin (i.e. "Principles of mathematical analysis") for Theorem 3.37, shown below.

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 $n \geq N$. In particular, for any $p > 0$, $$c_{N+k+1} \leq \beta c_{N+k}.$$ 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 $$\limsup_{n \to \infty} \sqrt[n]{c_n} \leq \alpha.$$

I am struggling to understand the steps that are taken in the following part of the proof:

Multiplying these inequalities, we obtain $$c_{N+p} \leq \beta^p c_N,$$ or $$c_n \leq c_N \beta^{-N} \cdot \beta^n \quad (n \geq N).$$

Specifically, I'm wondering:

  1. Where does the inequality $c_{N+p} \leq \beta^p c_N$ come from? Is the author saying that this is the result of multiplying the two preceding inequalities together, like $(\frac{c_{n+1}}{c_n} \cdot c_{N+k+1}) \leq (\beta \cdot \beta c_{N+k})$?
  2. What is the connection between $c_{N+p} \leq \beta^p c_N$ and the next inequality $c_n \leq c_N \beta^{-N} \beta^n$?

Any help would be very much appreciated!


Side note: I found this question to be really helpful with proving the first inequality.

$\endgroup$
3
2
$\begingroup$

We have $$c_{N+1}\le \beta c_{N}$$ $$c_{N+2} \le \beta c_{N+1} \le \beta(\beta c_N)=\beta^2 c_N$$

We can repeatedly do this and conclude that

$$c_{N+p}\le \beta^pc_N$$

For a more formal proof, use mathematical induction.

The next line is just a re-indexing, let $n=N+p$.

$$c_n \le \beta^{n-N}c_N= \beta^{-N}\cdot \beta^n c_N$$

$\endgroup$
1
$\begingroup$

(1) You write $p$ inequalities corresponding to the values $k=0$ to $k=p-1$ and multiply these inequalities.

(2) If $n \geq N$ we can write $n=N+p$ for some $p \geq 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.