Convergence of the Sequence \sqrt[n]{n} to 1

[Dodobird]

Homework Statement

Be $K \geq 1$. Conclude out of the statement that $\lim_{n \to \infty }$ $\sqrt[n]{n} = 1$, dass $\sqrt[n]{K} = 1$

The Attempt at a Solution

$\lim_{n \to \infty } \sqrt[n]{K} \Rightarrow 1 \leq \sqrt[n]{K} \geq 1 + ...$

I got issues with the right inequality, where the 3 dots are. I´m not sure if just insert the $\sqrt[n]{n}$ there and that s about it.

Thanks in advance ;)

Christian...

 

[Fightfish]

Let $1 \leq K \leq n$ for some big n (we're going to let it tend to infinity later)
Then we can do a squeeze:
$$\lim_{n\to\infty}\sqrt[n]{1} \leq \lim_{n\to\infty}\sqrt[n]{K} \leq \lim_{n\to\infty}\sqrt[n]{n}$$

 

[Dodobird]

Many thanks Fightfish for your quick reply so i can state the following:

For the 3 real Sequences $\sqrt[n]{1} , \sqrt[n]{K} , \sqrt[n]{n}$ $\exists N$ $\forall n \geq N$ is $\sqrt[n]{1} \leq \sqrt[n]{K} \leq \sqrt[n]{n}$

and $\lim_{n\to\infty}\sqrt[n]{1} = \lim_{n\to\infty}\sqrt[n]{n}$.


$\Rightarrow \sqrt[n]{K}$ converges and $\lim_{n\to\infty}\sqrt[n]{K} = \lim_{n\to\infty}\sqrt[n]{n}$

 

[Dodobird]

I thought about this and maybe this one here is more elegant than the other one, would be cool if someone could backcheck it.

$\forall n \text{ with } n > K >1: \sqrt[n]{K} <\sqrt[n]{n}$

and $\lim_{n\to\infty}\sqrt[n]{1}= 1$

$\lim_{n\to\infty}\sqrt[n]{K} \leq \lim_{n\to\infty}\sqrt[n]{n} = 1$

Thanks again for your helping...

 

[SammyS]

Homework Statement

Be $K \geq 1$. Conclude out of the statement that $\lim_{n \to \infty }$ $\sqrt[n]{n} = 1$, dass $\sqrt[n]{K} = 1$

The Attempt at a Solution

$\lim_{n \to \infty } \sqrt[n]{K} \Rightarrow 1 \leq \sqrt[n]{K} \geq 1 + ...$

I got issues with the right inequality, where the 3 dots are. I´m not sure if just insert the $\sqrt[n]{n}$ there and that s about it.

Thanks in advance ;)

Christian...

Hello Christian (Dodobird). Welcome to PF !
I also have issues with the inequality:

$\displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \geq 1 + ...$

Why do you have both ≤ and ≥ in the same compound inequality?

 

[Dodobird]

Thank you Sammy for your warm welcome
Oh yeah, you are right. Both signs should point in the same direction. I mistakenly wrote it in the wrong way.Sorry about that.
So it should be:

$\displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \leq 1 + ...$


Thx ;)

 

[SammyS]

Thank you Sammy for your warm welcome
Oh yeah, you are right. Both signs should point in the same direction. I mistakenly wrote it in the wrong way.Sorry about that.
So it should be:

$\displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \leq 1 + ...$

Thx ;)

I'm just making sure that I understand this exercise.

You are to prove, for K≥1 that $\lim_{n \to \infty }\sqrt[n]{K} = 1\,,$ using the result that $\lim_{n \to \infty }\sqrt[n]{n}=1\ .$ Is that correct?

 

[Dodobird]

Yeah, that´s correct Sammy. Do you see any flaws?
I´m pretty new to proofs in Mathematics and still struggle with it and still feel a little bit insecure when I got to prove something.