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Show that $\lim_\limits{n \rightarrow \infty}\sqrt[n]{n}=1$ using Bernoulli inequality

I have a hint $\epsilon > 0$ and $n=n\cdot \epsilon/\epsilon$ and I need to show $$1\leq \sqrt[n]{n}\leq \frac{1}{\sqrt[n]{\epsilon}}(1+\epsilon)$$

What I have done so far is verifying the inequality on the left. Also $$\sqrt[n]{n}= \frac{1}{\sqrt[n]{\epsilon}}\sqrt[n]{\epsilon}\sqrt[n]{n}$$

Thanks for taking your time

Lorenzo B.
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RM777
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5 Answers5

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$(1+\frac1{\sqrt{n}})^n \ge 1+\frac{n}{\sqrt{n}} \gt n^{1/2} $.

Raising to the $2/n$ power, $n^{1/n} =(n^{1/2})^{2/n} \lt ((1+\frac1{\sqrt{n}})^n)^{2/n} = (1+\frac1{\sqrt{n}})^2 =1+\frac{2}{\sqrt{n}}+\frac1{n} \lt 1+\frac{3}{\sqrt{n}} \to 1 $.

marty cohen
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  • Why is $ 1+\frac{n}{\sqrt{n}} \gt n^{1/2} $? – RM777 Nov 19 '18 at 17:50
  • $\frac{n}{n^{\frac{1}{2}}}=\frac{n^{\frac{1}{2}}n^{\frac{1}{2}}}{n^{\frac{1}{2}}}=n^{\frac{1}{2}}$ I don't think this is what they had in mind, when they gave the hint, but i have to admit, that it is by far the shortest solution. – Martin Erhardt Nov 19 '18 at 19:53
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This is how you are taking advantage of Bernoulli in order to obtain the second inequality:

$\frac{(1+\epsilon)^n}{\epsilon}\overset{Bernoulli}{\geq} \frac{1+n\epsilon}{\epsilon} = \frac{1}{\epsilon}+n \geq n \Rightarrow\frac{(1+\epsilon)}{\epsilon^{\frac{1}{n}}} \geq n^{\frac{1}{n}}$

Note that: $$\lim_{\epsilon\rightarrow0}\lim_{n\rightarrow\infty}\frac{(1+\epsilon)}{\epsilon^{\frac{1}{n}}}=\lim_{\epsilon\rightarrow 0}\left (\lim_{n\rightarrow\infty}\left (\frac{1}{\epsilon^{\frac{1}{n}}}+\frac{\epsilon}{\epsilon^{\frac{1}{n}}}\right ) \right )=\lim_{\epsilon\rightarrow 0}\left (\lim_{n\rightarrow\infty}\frac{1}{\epsilon^{\frac{1}{n}}}+\lim_{n\rightarrow\infty}\frac{\epsilon}{\epsilon^{\frac{1}{n}}} \right)$$ $$=\lim_{\epsilon\rightarrow 0}(\lim_{n\rightarrow\infty}\epsilon^{-\frac{1}{n}}+\lim_{n\rightarrow\infty}\epsilon^{1-\frac{1}{n}})=\lim_{\epsilon\rightarrow0}(1+\epsilon)=1$$ $\lim_{n\rightarrow\infty}\epsilon^{-\frac{1}{n}}=1$ because of $\epsilon$ being a constant.

To conclude: $1=\lim_{\epsilon\rightarrow0}\lim_{n\rightarrow\infty}\frac{(1+\epsilon)}{\epsilon^{\frac{1}{n}}} \geq \lim_{\epsilon\rightarrow0}\lim_{n\rightarrow\infty}n^{\frac{1}{n}}=\lim_{n\rightarrow\infty}n^{\frac{1}{n}}$

Martin Erhardt
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  • Substitute $\lim_{\epsilon \rightarrow 0}$ with $\lim_{k \rightarrow \infty}$, $\epsilon$ with $k^{-1}$ and the proof would still be correct. Just in case you are not familiar with this notion your in lecture. – Martin Erhardt Nov 18 '18 at 21:59
  • why epsillon goes to $0$? – RM777 Nov 19 '18 at 18:01
  • Also the line where you say $\lim_{\epsilon\rightarrow0}\lim_{n\rightarrow\infty}\frac{(1+\epsilon)}{\epsilon^{\frac{1}{n}}}=\lim_{\epsilon\rightarrow 0}(\lim_{n\rightarrow\infty}\epsilon^{-\frac{1}{n}}+\lim_{n\rightarrow\infty}\epsilon^{1-\frac{1}{n}})=\lim_{\epsilon\rightarrow0}(1+\epsilon)=1 $ , how do you get first equality in particular $\lim_{n\rightarrow \infty} \epsilon^{1-\frac{1}{n}}$? – RM777 Nov 19 '18 at 18:09
  • Suppose I have already proved $\lim_{n\rightarrow \infty} \sqrt[n]a = 1 , a \in (0,1)$. Can this help to make it easier? – RM777 Nov 19 '18 at 18:35
  • Let me simply link you to math.stackexchange.com/a/488042/254056 . The cases $a \in (0,1).a \in [1,∞)$ are basically the same. I suppose you can just assume it is known even though the proof is not long. – Martin Erhardt Nov 19 '18 at 19:30
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You can prove it like that: $(\sqrt n)^{1/n}=1+k_n$ Where $k_n$ is some positive number depending on $n$

Then $\sqrt n =(1+k_n)^n \geq nk_n$-Bernoulli

So $k_n \leq \sqrt n/n =1/\sqrt n$

And then $1 \leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/\sqrt n + 1/n$

And you get what you need using squeeze theorem

Anton
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Option.

$x \ge 0.$

$(\star)$ $(1+x)^n \gt (n^2/4)x^2$.

$x=2/√n.$

$(1+2/√n)^n \gt n$

$1+2/√n \gt n^{1/n} \gt 1.$

Proof of $(\star).$

$(1+x)^n \ge n(n-1)x^2/2...\gt$

$(n^2/4)x^2,$ since for $n >2$:

$(n-1) > n/2$.

Peter Szilas
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As shown in this answer, using Bernoulli's Inequality, $\left(1+\frac1n\right)^{n+1}$ is decreasing.

Thus, for $n\ge1$, $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac1{n+1}\left(1+\frac1n\right)^{n+1}\\ &\le\frac4{n+1} \end{align} $$ Therefore, for $n\ge3$, $$ \begin{align} \frac{(n+1)^{\frac1{n+1}}}{n^{\frac1n}} &=\left(\frac{(n+1)^n}{n^{n+1}}\right)^{\frac1{n(n+1)}}\\ &\le\left(\frac4{n+1}\right)^{\frac1{n(n+1)}}\\[12pt] &\le1 \end{align} $$ Thus, for $n\ge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $\lim\limits_{n\to\infty}n^{1/n}$ exists and is no less than $1$.

Let $\alpha=\lim\limits_{n\to\infty}n^{1/n}$, then $$ \begin{align} \alpha &=\lim_{n\to\infty}n^{\frac1n}\\ &=\lim_{n\to\infty}(2n)^{\frac1{2n}}\\ &=\lim_{n\to\infty}2^{\frac1{2n}}\lim_{n\to\infty}n^{\frac1{2n}}\\[3pt] &=1\cdot\sqrt{\alpha} \end{align} $$ Since $\alpha\ge1$ and $\alpha=\sqrt{\alpha}$, we get $$ \begin{align} \lim_{n\to\infty}n^{\frac1n} &=\alpha\\ &=1 \end{align} $$

robjohn
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