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.