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This is supposed to be proven: $$\lim\limits_{n\to\infty}\sqrt [n]n=1$$

The sequence is monotone decreasing and has a lower bound of $1$. So $\epsilon=0$

With the Archimedean Property we get:

$$n_0\gt1+\dfrac2{\epsilon^2}$$

(What's that? $2$? epsilon squared? Where does this come from?)

The binomic theorem yields: $$n=\left(\sqrt[n]{n}\right)^n =\left(1+\left(\sqrt[n]n-1\right)\right)^n =\sum_{k=0}^n\binom{n}{k}\left(\sqrt[n]{n}-1\right)^k \geqslant\binom n2\left(\sqrt[n]{n}-1\right)^2 \\\,\\ =\dfrac{n(n-1)}2\left(\sqrt[n]{n}-1\right)^2$$

(how do I know that: $\left(\sqrt[n]{n}\right)^n=\left(1+\left(\sqrt[n]n-1\right)\right)^n$ ?, why is $k$ suddenly $2$?)

Then: $$\left(\sqrt[n]{n}-1\right)^2\leqslant\dfrac2{n-1}$$ and $$0\lt\sqrt[n]n-1\leqslant\sqrt{\dfrac2{n-1}}\leqslant\sqrt{\dfrac2{n_0-1}}\lt\sqrt{\epsilon^2}=\epsilon$$ (Again: What is all this and where does it come from?)

Ultimately: $$\left|\sqrt[n]n-1\right|=\sqrt[n]n-1\lt\epsilon$$ which is supposed to prove: $$\left(\sqrt[n]{n}\right)_{n\geqslant1}\longrightarrow1$$

Thank you in advance.

Martin Sleziak
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user3071205
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4 Answers4

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The proof first notes that for all $n$ we have $\sqrt[n] n \ge 1$, so - given $\epsilon > 0$ it suffices to find $n_0$ such that for all $n \ge n_0$ we have $\sqrt[n] n \le 1 + \epsilon$, as this implies $|\sqrt[n] n - 1| \le \epsilon$ for all $n \ge n_0$.

To prove this, we choose an $n_0$ such that $n_0 \ge 1 + \frac 2{\epsilon^2}$ (such an $n_0$ existis by the Archimedian property of the reals), we exactly this number is choosen will be more clearly later on.

We have, doing almost nothing (adding 0), that $$ \def\n{\sqrt[n] n} \n = \n + 1 - 1 = 1 + (\n - 1) $$ Taking $n$-th powers, this gives by the binomial theorem $$ n = \n^n = \sum_{k=0}^n \binom nk 1^{n-k} (\n-1)^k = \sum_{k=0}^n \binom nk (\n-1)^k $$ As $\n - 1 \ge 0$, all summands above are non-negative, therefore the sum is greater all equal all its summands, especially the second $$ n = \sum_{k=0}^n \binom nk (\n-1)^k \ge \binom n2 (\n - 1)^2 $$ So, by rewriting the last inequality in terms of $\n$, we have $$ n \ge \binom n2 (\n - 1)^2 \iff (\n-1)^2 \le \frac 2{n-1} \iff \n \le 1 + \sqrt{\frac 2{n-1}} $$ Now for $n \ge n_0$, this is lower or equal $1 + \sqrt{\frac 2{n_0-1}}$ and we want this to be lower or equal to $1 + \epsilon$ (which holds exactly iff $n_0 \ge 1 + \frac 2{\epsilon^2}$ which we exactly have chosen above.

Now we are done: Given $\epsilon > 0$ we have found an $n_0$ such that $$ 1 \le \n \le 1 + \epsilon, \quad\quad n \ge n_0 $$ This proves $\n \to 1$.

martini
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1

If you expand $x^{1/x}$ as a Taylor series for an infinitely large values of $x$, the development is :

$$1 + \left[\dfrac{\log(x)}x\right] + \dfrac{\left[\dfrac{\log(x)}x\right]^2}2 + \ldots $$

Taking into account the behaviour of $\log(x)$ compared to the behaviour of $x$, the limit is $1$ by positive values. You then have what martini clearly proved using a more elegant approach.

Claude Leibovici
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0

At first, notice that $\sqrt[n]{n}=e^{\log(n)/n}$ and thus, it's enough to find the limit of $\frac{\log(n)}{n}.$ You can also notice that for every $n\geq3$: $$\frac{1}{n}\leq\frac{\log(n)}{n}\leq\sqrt{\frac{1}{n}}.$$ The first part is easy: because of $n\geq3$, we obtain that $\log(n)\geq1$ and thus, $\frac{1}{n}\leq\frac{\log(n)}{n}$. For the second part, it is enough to prove that $\ \log(n)\leq\sqrt{n}\ \ \ \text{-or-}\ \ \ n\leq e^{\sqrt{n}}$, which does hold for $n=3$. Now the inductive step: let it be true for some $n\in\mathbb{N};\ \ n+1\leq e^{\sqrt{n}}+1\leq e^{\sqrt{n+1}}$ which is obvious true. Hence, $n\leq e^{\sqrt{n}}\ \ \text{-or-}\ \ \log(n)\leq \sqrt{n}$ and all is proved. Since $\lim\frac{1}{n}=0$ and $\lim\sqrt{\frac{1}{n}}=0$, it is enough to show that $\lim\frac{\log(n)}{n}=0$ and conclude that $\sqrt[n]{n}\longrightarrow e^0=1$.

Theodoros Balis
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Let $$ \sqrt[2n]{n}=1+d_n. $$ Clearly, $d_n>0$ and $$ \sqrt{n}=(1+d_n)^n\ge 1+nd_n>nd_n. $$ Hence $$ \frac{1}{\sqrt{n}}\ge d_n>0, $$ which implies that $d_n\to 0$, and therefore $$ \sqrt[n]{n}=(1+d_n)^2\to 1. $$

Yiorgos S. Smyrlis
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