# Understanding Baby Rudin Theorem 3.37

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.

• Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. Scanned pages from books are discouraged on SE network. Questions should contain sufficient context so that it is answerable with the text alone. May 5 '19 at 10:08
• @GNUSupporter8964民主女神地下教會 I've edited the question to replace the image with text. May 12 '19 at 1:06
• Thanks for edit. I've also edited your question so as to put your list of questions under an HTML <ol> tag. Please consider the tag real-analysis for analysis on $\Bbb{R}$. May 12 '19 at 7:20

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 $$n=N+p$$.

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

(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 $$p \geq 0$$.