Higher Roots of Positive Numbers

[dom_quixote]

Playing around with my calculator, I realized that if I do successive rooting operations on any positive non-zero number, I always get the number one.
Can I conclude that the infinite root of any positive number will always be zero?
If the statement is true, is there any synthesized formula to prove this property?

P.S.:
Sorry if my way of expressing myself sounds strange, because I'm Brazilian and I use the automatic translator to communicate with all of you. It is a pleasure to observe the interest of the contributors to this Site, who strive to understand the questions I post on the Physics Forums.

 

[mfb]

$$\lim_{n\to \infty} a^{1/n} = 1$$ for all positive a. You can show this e.g. using $a^{1/n} = e^{\log(a)/n}$ and continuity of the exponential function to bring the limit into the exponent, but you can also show it for successive square roots in more elementary ways by showing that e.g. the distance to 1 decreases at least by a factor 2 each time.

 

[fresh_42]

Playing around with my calculator, I realized that if I do successive rooting operations on any positive non-zero number, I always get the number one.
Can I conclude that the infinite root of any positive number will always be zero?

No, you cannot conclude anything from trying out a couple of numbers. And there is no such thing like an infinite root. However, you can pose the hypothesis that
$$
\lim_{n \to \infty} \sqrt[n]{x} = 1 \text{ for }x>0
$$

If the statement is true, is there any synthesized formula to prove this property?

You can define a sequence $a_n:=\sqrt[n]{x}$ and prove that $|a_n-1|$ get's arbitrary small for large $n$.

 

[PeroK]

$$\lim_{n\to \infty} a^{1/n} = 1$$ for all positive a. You can show this e.g. using $a^{1/n} = e^{\log(a)/n}$ and continuity of the exponential function to bring the limit into the exponent, but you can also show it for successive square roots in more elementary ways by showing that e.g. the distance to 1 decreases at least by a factor 2 each time.

You can also do an indirect proof. For any $a > 1$, the sequence $a^{1/n}$ is monotone decreasing and bounded below by $1$. Therefore, the limit exists.

However, for any number $l > 1$, the sequence $l^n$ in unbounded; and so $l \ne lim_{n \to \infty} a^{1/n}$. That leaves the only possibility that $lim_{n \to \infty} a^{1/n} = 1$.