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

  • $\begingroup$ 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

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


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