The $n$-th root of $2$ is irrational for $n>1$, and $\lim_{n\to\infty}\sqrt[n]{2}=1$. Is the latter fact a corollary of the former?

Question

So I was looking at a theorem that popped up into my head:

The $n$-th root of $2$ always being irrational if $n$ is greater than $1$.

I also noticed that

The limit of the $n$-th root of $2$, as $n$ approaches infinity, is $1$.

Is there a connection between this "$n$-th root of $2$ Theorem" and the limit I mentioned?

Can I consider that limit a corollary to the theorem?

Answer 1

As far as I understand your question, no. The sequence

$$2\sqrt{2},2\root3\of2,2\root4\of2,\cdots$$

(multiplying every term of your nth-root-of-$2$ sequence by $2$) contains only irrational terms and tends to $2$, while the sequence

$$1+1,1+\frac{1}{2},1+\frac{1}{3},\cdots$$

contains only rational terms and tends to $1$. The two properties (containing irrational terms and tending to 1) have nothing to do with one another.

Answer 2

The two things are not related, for the limit indeed we have that

$$\sqrt[n] 2=e^{\frac{\log 2}n}\to 1$$

since $\frac{\log 2}n\to 0$, and the result is not related to any hypotesis about the irrationality of $\sqrt[n] 2$.

Answer 3

The $n^{th}$ root of any positive number tends toward $1$ as $n \to \infty$. This is regardless of whether the roots are rational or irrational. For any given positive number there will only be a finite number of rational roots, but it can be arbitrarily large. Find some $k$ with many factors, like one of the highly composite numbers. Now form $1.01^k$. That will have as many rational roots as $k$ has factors.