0

Here's my attempt at trying to prove it by definition:

$$\exists \mathcal{E} > 0 | \exists N\in \mathbb{N}|\forall n > N:$$ $$|\sqrt[n]{n}-1|<\mathcal{E}$$

But from this point on, I'm not sure how to demonstrate that $N$. I tried applying the mean inequality, but when I do, the indexes cancel out and I get something like $1< \mathcal{E}$, which negates the proof. How can I proceed? I also don't know how to utilise the fact that the infimum is equal to one.

daedsidog
  • 997
  • 1
    Since $\sqrt[n]n=1$ is $n=1$ and $\sqrt[n]n>1$ otherwise, it is trivial that $\inf\left{\sqrt[n]n\mid n\in\mathbb N\right}=1$. I don't see how is it that you expect to get some information from a trivial assertion. – José Carlos Santos Apr 18 '19 at 18:50
  • I don't think this is a duplicate. OP wants to prove a statement taking the result of the linked question as given. – Umberto P. Apr 18 '19 at 20:45
  • @UmbertoP. linked question contains sufficient material to do so, i.e. proving the $\inf$ and the $\lim$. – rtybase Apr 18 '19 at 20:49

2 Answers2

1

The limit is equivalent to showing that $\log (n) /n $ tends to 0, then exponentiating. Can you prove the second claim?

Amichai Lampert
  • 501
0

I think you meant to assume that $\inf\{\sqrt[n]{n} \mid n \ge 2\} = 1$.

The sequence $\{\sqrt[n]{n}\}$ is eventually decreasing because $$ \sqrt[n]{n} \ge \sqrt[n+1]{n+1} \iff n^{n+1} \ge (n+1)^n \iff n \ge \left( \frac{n+1}n \right)^n.$$ The terms on the left hand side of the last inequality diverge to $0$, but the terms on the right converge to $e$. Thus there exists an index $N$ with the property that $$ n \ge N \implies \sqrt[n]{n} \ge \sqrt[n+1]{n+1}.$$

It is (hopefully) obvious that $\sqrt[n]{n} > 1$ for all $n \ge 2$. Thus given any $\epsilon > 0$ there exists a index $N' \ge N$ with the property that $$ 1 < \sqrt[N']{N'} < 1 + \epsilon.$$

Consequently $$ n \ge N' \implies 1 < \sqrt[n]{n} < 1 + \epsilon.$$

Umberto P.
  • 52,165