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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.

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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.

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  • $\begingroup$ What do you mean by beta < alpha $\endgroup$
    – user264745
    Sep 18 at 19:27
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    $\begingroup$ The first sentence of your answer is a complete answer. The second sentence doesn't make any sense - $\endgroup$
    – Rob Arthan
    Sep 18 at 20:12
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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 $\lim_{n\to \infty} \sup \sqrt[n]{c_n}$

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