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This is the one:

$$\lim\limits_{n\to \infty}\sqrt[n]{n} = 1$$

With $\varepsilon > 0$ the Archimedean Property of Reals yields an $n_0 \in \mathbb{N}$ with

$$n_0 > 1 + \dfrac{2}{\varepsilon^2}$$

(I really don't get how the Archimedean Property yields this. I know it as "for each $a,b$ of $\mathbb{R}$ there is an $n$ such that $na>b$". Simple. But how does it yield the above?)

Then we have $n \in \mathbb{N}$ with $n \ge n_0$ and especially $ \ge 2$. The Binomial theorem yields:

$$\begin{align*}n = \left(\sqrt[n]{n}\right)^n = (1 + (\sqrt[n]{n} -1))^n &= \sum\limits_{k =0}^n {n \choose k} (\sqrt[n]{n}-1)^k \\&\ge {n \choose 2}(\sqrt[n]{n}-1)^2 \\&= \dfrac{n(n-1)}{2}\cdot \left(\sqrt[n]{n}-1\right)^2 \end{align*}$$

Hence:

$$(\sqrt[n]{n}-1)^2 \le \dfrac{2}{n-1} \quad \text{ and }\quad 0 < \sqrt[n]{n}-1\le \sqrt{\dfrac{2}{n-1}}\le \sqrt{\dfrac{2}{n_0-1}}< \sqrt{\varepsilon^2}=\varepsilon$$

I'm confused. How does this result from the step before?

And finally

$$\vert\sqrt[n]{n}-1\vert = \sqrt[n]{n}-1 < \varepsilon.$$

which is supposed to prove the limit. Again, I don't get how it is derived.

Bman72
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user3071205
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4 Answers4

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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

$n_0 > 1 +\dfrac{2}{\epsilon^2} \iff n_0 - 1 > \dfrac{2}{\epsilon^2} \iff \dfrac{1}{n-1} < \dfrac{\epsilon^2}{2} \iff \sqrt{\dfrac{2}{n-1}} < \epsilon$.

John Habert
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Hint: $n^{1/n} = \exp((\log n)/n)$. Now use L'Hopital's rule.

user901823
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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.

user10444
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  • How is the n0 = $1+\dfrac{2}{\epsilon^2}$ derived? – user3071205 Feb 01 '14 at 20:27
  • @user3071205 usually these are chosen after you do the proof. In the last step you pick which $n_0$ such that when $n>n_0$ the sequence and its limit are close. When writing the proof however the value is directly stated. If you were doing the proof and you got to showing $|x_n-1|<\sqrt{\frac {2}{n-1}}$ you would have picked such an $n_0$ – user10444 Feb 01 '14 at 20:32
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Clearly what you really want to know is how in the world someone came up with the idea to write this. I will break it down for you:

Firstly you want $n^{\frac{1}{n}} - 1 < \varepsilon$ which you rearrange to $n^{\frac{1}{n}} < \varepsilon + 1$. To simplify the left hand side so that you only have to deal with polynomials, which are often the best-understood objects in math, you take the $n$th power of both sides and expand out the right side by the Binomial theoerm to get $$n < 1 + n\varepsilon + \frac{n(n - 1)}{2}\varepsilon^2 + \dots$$ The easiest way to assert this is true would be if one of the terms on the right side were by itself greater than $n$. However, $1 > n$ and $n\varepsilon > n$ are impossible for $n > 1$ and $\varepsilon < 1$, so they are bad candidates. Thus, you try the next term, and find $$n < \frac{n(n - 1)}{2}\varepsilon^2 \Longleftrightarrow n > \frac{2}{\varepsilon^2} + 1$$ by some simple algebraic rearrangement, and then you just finish the details of the proof/write it out.

As for the Archimedean property, instead of thinking of the formal statement, try to think of it intuitively as the natural fact that every real number is between two consecutive integers, which is one of the equivalent forms to state it. This is essentially saying there are no "infinite" real numbers, and no "infinitesimal" reals either. See https://en.wikipedia.org/wiki/Hyperreal_number for an example of why you might want to specify this nice property of the real numbers, which seems so naturally obvious to begin with.

Once you have an intuition, proving the equivalence of the various formulations of the Archimedean property is a good exercise in formality.

MathNeophyte
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