Given $a>0$ prove that $\lim a^{1/n} =1$. One of my tries was to find a number $N \in R / |a^{1/n}-1|< \epsilon, \forall n\ge N$so I started by $|a^{1/N}-1|< \epsilon$ then separated the inequality for $a>1$and then I got $a^{1/N}-1<\epsilon$, then $1/N<log_a(\epsilon -1)$, then I was trying to find a way to make sure that $log_a(\epsilon -1)> 0$ but I couldn't.
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$\begingroup$For $a=1$ the claim holds, because constant sequence will have the same limit. Assume $a>1$, then $a^{1/n} > 1$ (why?). Put $x_n := a^{1/n} -1 > 0, n\in\mathbb N$. By the binomial theorem$$a = (1+x_n)^n > nx_n \implies 0 < x_n < \frac{a}{n}\xrightarrow[n\to\infty]{}0.$$For $0<a<1$ put $b := \frac{1}{a} >1$ and $$\sqrt[n]{a} = \sqrt[n]{\frac{1}{b}} = \frac{1}{\sqrt[n]{b}} \xrightarrow[n\to\infty]{}1. $$
Alternatively, as suggested by @herb steinberg, we could use$$ \frac{\ln a}{n} \xrightarrow[n\to\infty]{}0\qquad (a>0)$$which implies due to continuity of the exponential function$$ a^{1/n} = e^{\ln a/n} \xrightarrow[n\to\infty]{}e^0 = 1. $$
$\endgroup$ $\begingroup$There is a nice proof involving the AM-GM inequality and the squeeze theorem.
Let $a\ge 1$. Write $a=\sqrt{a}\cdot\sqrt{a}\cdot 1\cdots 1$ (with $n-2$ ones). By the AM-GM we have$$1\le a^{1/n}=(\sqrt{a}\cdot\sqrt{a}\cdot 1\cdots 1)^{1/n}\le \frac{2\sqrt{a}+n-2}{n}\to 1.$$The squeeze theorem concludes the proof.
The case $0<a<1$ is the immediate consequence of the case $a\ge 1$ (use $1/a$ instead of $a$).
Remark. In the same way one can show that $\sqrt[n]{n}\to 1$.
$\endgroup$ 0 $\begingroup$Show:$$\left|a^{1/n}-1\right|=\frac{\left|a-1\right|}{1+a^{1/n}+a^{2/n}+\cdots+a^{(n-1)/n}}\leq \frac{|a-1|}{n\min(1,a)}$$
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