Prove limit of n-th root This is the one:

With epsilon > 0 the Archimedean Property of Reals yields an  with

(I really don't get how the Archimedean Property yields this. I know it as "for each a,b of R there is an n such that na>b". Simple. But how does it yield the above?) 
Then we have  with  and especially n>=2 (Why?). Binomial theorem yields:

Hence:
 and 
I'm confused. How does this result from the step before?
And finally

which is supposed to prove the limit. Again, I don't get how it is derived.
 A: Using your version of the Archimedean Principle, let $a=1$ and $b=1+\dfrac{2}{\epsilon^2}$. Then you get that inequality.

You need especially $n\geq2$ since if $n=0,1$ it is not true that $\displaystyle \sum_{k=0}^n \binom{n}{k}(\sqrt[n]{n}-1)^k \geq \binom{n}{2}(\sqrt[n]{n}-1)^2$.

The Binomial Theorem yields $n \geq \dfrac{n(n-1)}{2}(\sqrt[n]{n}-1)^2$ which can be rearranged into $(\sqrt[n]{n}-1)^2 \leq \dfrac{2}{n-1}$.

Going back to the choice of $n_0$, we see that
$n_0 > 1 +\dfrac{2}{\epsilon^2} \iff n_0 - 1 > \dfrac{2}{\epsilon^2} \iff \dfrac{1}{n-1} < \dfrac{\epsilon^2}{2} \iff \sqrt{\dfrac{2}{n-1}} < \epsilon$.
A: Hint: $n^{1/n} = \exp((\log n)/n)$. Now use L'Hopital's rule.
A: Notice that since $n_0-1 \in \Bbb N$ then the property gives us  for some $\epsilon^2$ ,$\epsilon^2(n_0-1)>2$, this gives $\epsilon^2n_0-\epsilon^2>2 $ then $\epsilon^2n_0>2+\epsilon^2$ then divide $\epsilon^2$ to get the inequality.
From the whole binomial expansion we are only taking the second term hence you get $\sum_{k=0}^n {n \choose  k }(\sqrt[ n]{n}-1)^k \ge {n \choose  2 }(\sqrt[ n]{n}-1)^2 $. For large enough $n$ in this case bigger than $2$.
You now have that $n> {n \choose  2 }(\sqrt[ n]{n}-1)^2= \frac {n(n-1)}{2}(\sqrt[n]{n} -1)^2$ devide both sides be $\frac {n(n-1)}{2}$ to get $(\sqrt[n]{n} -1)^2 \le \frac 2 {n-1}$.
If I missed something you didn't understand, comment.
