# Theorem 3.37 of Baby Rudin

In theorem 3.37 of Baby Rudin. The following claim I can’t understand. $$\lim_{n \to \infty} \sup \sqrt[n]{c_{n}} \le \beta$$ is true $$\forall \beta \ge \alpha$$ then we have $$\lim_{n \to \infty} \sup \sqrt[n]{c_n} \le \alpha$$. I think it related with the concept of supremum, but still I can not prove it satisfactory.

Given inequality holds for all $$\beta \geq \alpha$$ so just take $$\beta = \alpha$$. If however $$\beta < \alpha$$ then u can how it using basic properties of limits.

• What do you mean by beta < alpha Sep 18 at 19:27
• The first sentence of your answer is a complete answer. The second sentence doesn't make any sense - Sep 18 at 20:12

This actually has nothing to do sequences and limits or supremema. The following it true for any real numbers no matter what the context.

If $$M \le \beta, \forall \beta \ge \alpha$$ then $$M \le \alpha$$

Pf: If $$M \le \beta$$ for all $$\beta \ge \alpha$$, then for every $$\beta \ge \alpha$$ we always have $$M \le \alpha \le \beta$$. So..... $$M \le \alpha$$.

That's all it's saying. It isn't deep. It was probably assumed to be obvious.

And if $$M = \lim_{n\to \infty}\sup \sqrt[n]{c_n}$$ then it is true for $$\lim_{n\to \infty} \sup \sqrt[n]{c_n}$$