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Part (b) of the theorem states that "If $p>0$, then $\lim_{n\to\infty}p^{1/n}=1.$"

Here is the $p\ge1$ case that Rudin proves:

If $p\ge1$, put $x_n=p^{1/n}-1\ge0.$ By the binomial theorem we have $1+nx_n\le(1+x_n)^n=p.$ We then have$$0\le x_n\le \frac{p-1}{n}$$ and $x_n\to 0.$

He then states that in the case for $0<p<1$, "the result is obtained by taking reciprocals." But he is using Bernoulli's inequallity and that hold for $x_n\ge -1$ and I just can't figure out how taking the reciprocal is useful. I'm certain that I am just being very thick headed but can someone spell out what he means?

Philip Shen
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    If $0<p<1$ then $p=\frac{1}{q}$ where $q>1$ so $p^{1/n}=\frac{1}{q^{1/n}}\to 1$ since the statement is already proved for $q>1$ – Marios Gretsas Sep 18 '21 at 05:11
  • Thanks @Marios Gretsas – Philip Shen Sep 18 '21 at 05:12
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    For what it's worth, if logarithms are allowed, the solution is easier. For all $\displaystyle x \in \Bbb{R}, ~\lim_{n \to \infty} \frac{x}{n} = 0.$ This holds, regardless of whether $x < 0, x = 0$ or $x > 0.$ Then $\displaystyle \log\left[p^{(1/n)}\right] = \frac{\log(p)}{n} ~: ~p > 0.$ So, as $\displaystyle n \to \infty, \log\left[p^{(1/n)}\right] ~: ~p > 0$ goes to $0$. This implies that $p^{(1/n)} ~: ~p > 0$ goes to $1$. – user2661923 Sep 18 '21 at 06:19
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    Use theorem 3.3 (d) to get your desired result. – JRB Dec 04 '21 at 11:32

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