Understanding Baby Rudin Theorem 3.37

Question

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!

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

Answer 1

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 $n=N+p$.

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

Answer 2

(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 $p \geq 0$.