# alternative proof for limit

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

• I don’t understand the first step. – gimusi Jan 28 '18 at 15:29
• 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. – user228113 Jan 28 '18 at 15:31
• 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} – user370967 Jan 28 '18 at 15:48
• thank you for help, i think now i should be right. – sabi Jan 28 '18 at 17:01
• math.stackexchange.com/questions/115822/… – Hans Lundmark Jan 28 '18 at 17:45

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