-1
$\begingroup$

Can someone check my proof for $\lim_{n \rightarrow\infty}\sqrt[n] n =1$ and show me maby an alternative proof ?

My proof: Be $\epsilon > 0 $ arbitrary, i know that $\lim_{n \rightarrow \infty} n^kx^n = 0$ for $ k \in \mathbb{N}$ and $x \in \mathbb{C}$ with $\vert x \vert < 1$. Set $x:=\frac{1}{1+ \epsilon}$ and $k=1$.

Then i get,

$\lim_{n \rightarrow\infty}n (1+ \epsilon)^{-n}=0 \Longrightarrow n(1+ \epsilon)^{-n}<1 $ for almoast all $n \in \mathbb{N}$.

That implies $1 \le n<(1+\epsilon)^{n} \Longrightarrow 1\le \sqrt[n]n<1+\epsilon$

So i get

$1 \le \lim_{n\rightarrow \infty} \sqrt[n]n=1+\epsilon \; \;\forall \epsilon >0$

$\epsilon $ was arbitrary so it follows that $$\lim_{n \rightarrow \infty} \sqrt[n]n=1$$

$\endgroup$
  • $\begingroup$ I don’t understand the first step. $\endgroup$ – gimusi Jan 28 '18 at 15:29
  • $\begingroup$ I think you had a sensible idea which got lost in a considerable amount of material errors in writing the algebra; specifically, in the third and fourth lines. Please, read again what you've written and correct it. $\endgroup$ – user228113 Jan 28 '18 at 15:31
  • $\begingroup$ You have to be more careful. If this was a good prood (which it isn't), it would have shown that 1 < lim n^{1/n} $\endgroup$ – user370967 Jan 28 '18 at 15:48
  • $\begingroup$ thank you for help, i think now i should be right. $\endgroup$ – sabi Jan 28 '18 at 17:01
  • $\begingroup$ math.stackexchange.com/questions/115822/… $\endgroup$ – Hans Lundmark Jan 28 '18 at 17:45
2
$\begingroup$

Alternatively you can proof it like follows:

$\sqrt[n]n=1+h_n$ for one $h_n \ge0$, powering the equality by $n$ you receive :

$$n=(1+h_n)^n = \sum_{i=0}^{n} \binom{n}{i}h_{n}^{i} \ge 1 + \frac{n(n-1)}{2} \cdot h_{n}^{2} $$ By subtracting $-1$ on both sides you get

$$n-1 \ge \frac{n(n-1)}{2} \cdot h_n^2$$

That implies with $\frac{2}{n} \ge h_n^2 \ge 0 $ that $h_n^2 \longrightarrow0$ and therefore $h_n \longrightarrow0$

Finally you get: $$\lim_{n \rightarrow \infty}\sqrt[n]n =\lim_{n \rightarrow \infty}(1+h_n)=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.