Infimum of $n$-th root of $n$

Question

Let $A = \left \{ \sqrt[n]{n} \mid n \in \mathbb{N}\right \}$. I need to find and prove the infimum of $A$.

Because $n \in \mathbb{N},$ we can know for sure that $\sqrt[n]{n} \geq 1$. (Does that need a deeper proof?)

Then to prove that $1$ is the infimum, for $\forall \varepsilon>0$, I need to find a certain $n \in \mathbb{N}$ that makes $$\sqrt[n]{n} < 1 + \varepsilon$$

How do I "engineer" that $n \in \mathbb{N}$ ?

What's the trick? Raising $\sqrt[n]{n} < 1 + \varepsilon $ to the power of $n$ doesn't really help.

Are you allowed to use the fact that $\lim\limits_{n\rightarrow\infty} \sqrt[n]{n}=1$? — rtybase, Feb 07 '19 at 11:31

José Carlos Santos · Answer 1 · 2019-02-07T11:31:56.620

1

The infimum of $A$ is $1$ because $1\in A$ (just take $n=1$) and $n>1\implies\sqrt[n]n>1$.

On the other hand, you can write $\sqrt[n]n$ as $1+\varepsilon_n$ and then\begin{align}n&=\sqrt[n]n^n\\&=(1+\varepsilon_n)^n\\&=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots\\&>\frac{n(n-1)}2{\varepsilon_n}^2\end{align}and therefore$$\varepsilon_n<\sqrt{\frac2{n-1}},$$if $n>1$.

edited Feb 07 '19 at 11:31

answered Feb 07 '19 at 11:17

José Carlos Santos

427,504

That's not enough. I need to prove that for each $ \varepsilon > 0 $ there is $ n \in \mathbb{N} $ that makes $n > $ some epsilon that we render.. – eran-avrahami Feb 07 '19 at 11:27
What do you think now? – José Carlos Santos Feb 07 '19 at 11:32
Well, it seems better but could I please get an explanation do what you've done? – eran-avrahami Feb 07 '19 at 11:43
@Eran19978 I am sorry, but I don't see what is there to explain. Could you be more specific? – José Carlos Santos Feb 07 '19 at 12:09
2

If you keep the $1$ and only drop the $n\epsilon_n$, you get the slightly better inequality $\epsilon_n\lt\sqrt{2/n}$ for $n\gt1$ (by canceling an $n-1$ instead of an $n$). – Barry Cipra Feb 07 '19 at 12:17
@BarryCipra Nice observation! – José Carlos Santos Feb 07 '19 at 12:22
@JoséCarlosSantos, I understand that you developed $ (1+\varepsilon)^{n} $ into its "binomial" form if you will, but why did you decide to "dig" $\frac{n(n-1)}{2}*\varepsilon{^2}$ ? – eran-avrahami Feb 07 '19 at 12:36
Because it works. I could have used$$\frac{n(n-1)(n-2)}6{\varepsilon_n}^3$$instead and it would have worked too, but why would I do that when $\frac{n(n-1)}2{\varepsilon_n}^2$ is enough? – José Carlos Santos Feb 07 '19 at 12:39
@JoséCarlosSantos To be more specific, I understand why $ (1+\varepsilon)^{n} > \frac{n(n-1)}{2}*\varepsilon{^2} $
But how do you determine that $ \frac{n(n-1)}{2}*\varepsilon{^2} > n $ ?
– eran-avrahami Feb 07 '19 at 12:40
Because$$n=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots$$and all terms in this sum are greater than $0$. Therefore, $n$ is greater than each of these terms. – José Carlos Santos Feb 07 '19 at 12:46
@JoséCarlosSantos Oh, I didn't see that you defined $\sqrt[n]n$ to be $ 1+\varepsilon_n $
What allows you to do that?
– eran-avrahami Feb 07 '19 at 12:51
@JoséCarlosSantos its kinda odd, because I need to show $ \sqrt[n]{n} < 1 + \varepsilon $
Then why would you define $ \sqrt[n]{n} = 1 + \varepsilon $
– eran-avrahami Feb 07 '19 at 12:54
I dont understand your doubt. I wrote $\sqrt[n]n$ as $1+\varepsilon_n$. In other words, I defined $\varepsilon_n$ as $\sqrt[n]n-1$. – José Carlos Santos Feb 07 '19 at 13:03
@JoséCarlosSantos Last question: Could I just Assume that if i find n that makes: $ n< \frac{n(n-1)}{2}*\varepsilon{^2}$ that would automatically apply for $n<(1+\varepsilon){^n}$ – eran-avrahami Feb 07 '19 at 14:37
1

Yes, since $\frac{n(n-1)}2\varepsilon^2\leqslant(1+\varepsilon)^n$. – José Carlos Santos Feb 07 '19 at 14:55

rtybase · Answer 2 · 2019-02-07T12:34:00.077

If you are not allowed to use the fact that $\lim\limits_{n\rightarrow\infty} \sqrt[n]{n}=1$ (which immediately leads to $\sqrt[n]{n} < 1+\varepsilon$ from the definition of limit), then try the following trick ...

$$n=e^{\ln{n}} \Rightarrow \sqrt[n]{n}=e^{\frac{\ln{n}}{n}}$$ It is not too difficult to show that $0<\frac{\ln{n}}{n} < 1$ and thus, we can apply Bernoulli's inequality $$e^{\frac{\ln{n}}{n}}=(1+e-1)^{\frac{\ln{n}}{n}}\leq 1+\frac{(e-1)\ln{n}}{n}$$ or $$\sqrt[n]{n}\leq 1+\frac{(e-1)\ln{n}}{n}$$ and because $\frac{n}{\ln{n}}>\sqrt{n}$ for $n>1$ we have $$\sqrt[n]{n}\leq 1+\frac{(e-1)\ln{n}}{n}<1+\frac{e-1}{\sqrt{n}}$$ you can squeeze $\frac{e-1}{\sqrt{n}} < \varepsilon$ now.

Now, why

$\frac{n}{\ln{n}}>\sqrt{n}$ for $n>1$

Because, using induction $$\frac{2}{\ln{2}} > \sqrt{2},\space \frac{3}{\ln{3}} > \sqrt{3},\space \frac{4}{\ln{4}} > \sqrt{4}$$ and $$\frac{n}{\ln{n}}>\sqrt{n} \iff \color{red}{\sqrt{n} > \ln{n}} \iff \sqrt{n} + \ln{\left(1+\frac{1}{n}\right)} > \ln{n} + \ln{\left(1+\frac{1}{n}\right)}=\ln{(n+1)}$$ using $\ln{(1+x)}<x$ from here $$\color{red}{\sqrt{n+1}>}\sqrt{n} +\frac{1}{n}>\sqrt{n} + \ln{\left(1+\frac{1}{n}\right)} >\color{red}{\ln{(n+1)}}$$ for $n\geq 5$

Infimum of $n$-th root of $n$

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