Finding limit of (2n)^(1/n) where n is natural number

[issacnewton]

HelloI am trying to prove that $\lim\left((2n)^{1/n}\right) = 1$.
Here $n\in \mathbb{N}$. I have already proven that $(2n)^{1/n} > 1$
for $n > 1$. So we can write $(2n)^{1/n} = 1 + k$ for some $k > 0$ when
$n>1$. Hence $2n = (1+k)^n$ for $n>1$. By the Binomial theorem, if $n>1$, we have,
\[ 2n = 1+nk+ \frac{1}{2}n(n-1)k^2+\cdots + k^n \]
Because all terms in Binomial expansion are positive, we can write
\[ 2n \geqslant 1 + \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow \; 2n > \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow 4 > (n-1) k^2 \]
\[ \Rightarrow k^2 < \frac{4}{n-1} \]
Above is true if $n>1$.
\[ \Rightarrow k^2 < \left(\frac{2}{\sqrt{n-1}}\right)^2 \cdots\cdots (1)\]
Now I use the following theorem I have already proven. Given $a\geqslant 0 $,
and $b\geqslant 0 $, then
\[ a<b \Longleftrightarrow a^2 < b^2 \Longleftrightarrow \sqrt{a}< \sqrt{b} \]
Now $n>1$, hence $(n-1) > 0$. Applying above theorem with $a=n-1$ and
$b=0$, we arrive at the conclusion that $\sqrt{n-1} > 0$. Hence $\frac{1}{\sqrt{n-1}} > 0$. Hence $\frac{2}{\sqrt{n-1}} > 0$.
Since $k>0$, again applying the above
stated theorem to the inequality $(1)$, we get
\[ k < \frac{2}{\sqrt{n-1}} \cdots\cdots (2)\]
for $n>1$. Now if we are given that $\varepsilon > 0$ is arbitrary, then $\frac{4}{\varepsilon^2}>0$.
\[1+\frac{4}{\varepsilon^2} > 1\]
By Archimedean Property, there exists a natural number $N_1$ such that
\[ 1< 1 + \frac{4}{\varepsilon^2} < N_1\]
\[\Rightarrow 0 < \frac{4}{\varepsilon^2} < N_1 - 1\]
\[ \Rightarrow \frac{\varepsilon^2}{4} > \frac{1}{N_1 - 1} > 0 \]
\[ \Rightarrow \left( \frac{\varepsilon}{2}\right)^2 > \left(\frac{1}{\sqrt{N_1 - 1}}\right)^2 \cdots\cdots(3)\]
Now $\frac{\varepsilon}{2} > 0$ and $N_1 >1$, so $N_1 - 1 > 0$. By above mentioned theorem, we have
$\sqrt{N_1 - 1} > 0$ , which implies $\frac{1}{\sqrt{N_1 - 1}}>0$. So applying the same theorem about the
square roots of nonnegative numbers to the equation $(3)$, we arrive at the conclusion that
\[\frac{\varepsilon}{2} > \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{N_1 - 1}} < \varepsilon \cdots\cdots(4)\]
Now for all $n \geqslant N_1$, because $N_1 > 1$, we have $n \geqslant N_1 > 1$.
\[ \Rightarrow n-1 \geqslant N_1 - 1 > 0 \]
\[ 0 < \frac{1}{n-1} \leqslant \frac{1}{N_1 - 1} \]
Again applying the same theorem, we get
\[ \frac{1}{\sqrt{n-1}} \leqslant \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{n-1}} \leqslant \frac{2}{\sqrt{N_1 - 1}} \]
Combining with equation $(2)$ and equation $(4)$, we conclude that, for all $n\geqslant N_1$,
$k < \varepsilon$. But since $(2n)^{1/n} = 1 + k$ , and $k > 0$, we see that
\[ | (2n)^{1/n} - 1 | = k < \varepsilon \;\;\; \forall n \geqslant N_1 \]
Since $\varepsilon$ is arbitrary, this proves that $\lim\left((2n)^{1/n}\right) = 1$.Please let me know if proof sounds right.thanks

 

[zzephod]

HelloI am trying to prove that $\lim\left((2n)^{1/n}\right) = 1$.
Here $n\in \mathbb{N}$. I have already proven that $(2n)^{1/n} > 1$
for $n > 1$. So we can write $(2n)^{1/n} = 1 + k$ for some $k > 0$ when
$n>1$. Hence $2n = (1+k)^n$ for $n>1$. By the Binomial theorem, if $n>1$, we have,
\[ 2n = 1+nk+ \frac{1}{2}n(n-1)k^2+\cdots + k^n \]
Because all terms in Binomial expansion are positive, we can write
\[ 2n \geqslant 1 + \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow \; 2n > \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow 4 > (n-1) k^2 \]
\[ \Rightarrow k^2 < \frac{4}{n-1} \]
Above is true if $n>1$.
\[ \Rightarrow k^2 < \left(\frac{2}{\sqrt{n-1}}\right)^2 \cdots\cdots (1)\]
Now I use the following theorem I have already proven. Given $a\geqslant 0 $,
and $b\geqslant 0 $, then
\[ a<b \Longleftrightarrow a^2 < b^2 \Longleftrightarrow \sqrt{a}< \sqrt{b} \]
Now $n>1$, hence $(n-1) > 0$. Applying above theorem with $a=n-1$ and
$b=0$, we arrive at the conclusion that $\sqrt{n-1} > 0$. Hence $\frac{1}{\sqrt{n-1}} > 0$. Hence $\frac{2}{\sqrt{n-1}} > 0$.
Since $k>0$, again applying the above
stated theorem to the inequality $(1)$, we get
\[ k < \frac{2}{\sqrt{n-1}} \cdots\cdots (2)\]
for $n>1$. Now if we are given that $\varepsilon > 0$ is arbitrary, then $\frac{4}{\varepsilon^2}>0$.
\[1+\frac{4}{\varepsilon^2} > 1\]
By Archimedean Property, there exists a natural number $N_1$ such that
\[ 1< 1 + \frac{4}{\varepsilon^2} < N_1\]
\[\Rightarrow 0 < \frac{4}{\varepsilon^2} < N_1 - 1\]
\[ \Rightarrow \frac{\varepsilon^2}{4} > \frac{1}{N_1 - 1} > 0 \]
\[ \Rightarrow \left( \frac{\varepsilon}{2}\right)^2 > \left(\frac{1}{\sqrt{N_1 - 1}}\right)^2 \cdots\cdots(3)\]
Now $\frac{\varepsilon}{2} > 0$ and $N_1 >1$, so $N_1 - 1 > 0$. By above mentioned theorem, we have
$\sqrt{N_1 - 1} > 0$ , which implies $\frac{1}{\sqrt{N_1 - 1}}>0$. So applying the same theorem about the
square roots of nonnegative numbers to the equation $(3)$, we arrive at the conclusion that
\[\frac{\varepsilon}{2} > \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{N_1 - 1}} < \varepsilon \cdots\cdots(4)\]
Now for all $n \geqslant N_1$, because $N_1 > 1$, we have $n \geqslant N_1 > 1$.
\[ \Rightarrow n-1 \geqslant N_1 - 1 > 0 \]
\[ 0 < \frac{1}{n-1} \leqslant \frac{1}{N_1 - 1} \]
Again applying the same theorem, we get
\[ \frac{1}{\sqrt{n-1}} \leqslant \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{n-1}} \leqslant \frac{2}{\sqrt{N_1 - 1}} \]
Combining with equation $(2)$ and equation $(4)$, we conclude that, for all $n\geqslant N_1$,
$k < \varepsilon$. But since $(2n)^{1/n} = 1 + k$ , and $k > 0$, we see that
\[ | (2n)^{1/n} - 1 | = k < \varepsilon \;\;\; \forall n \geqslant N_1 \]
Since $\varepsilon$ is arbitrary, this proves that $\lim\left((2n)^{1/n}\right) = 1$.Please let me know if proof sounds right.thanks

Prove that $$\lim_{n \to \infty} \ln( (2n)^{1/n} ) =0$$, you should only need L'Hopital's rule.

.

 

[issacnewton]

thanks for input zzephod. But is my present work correct ?

 

[Prove It]

Prove that $$\lim_{n \to \infty} \ln( (2n)^{1/n} ) =0$$, you should only need L'Hopital's rule.

.

Except the limit is actually 1, not 0.

 

[Deveno]

HelloI am trying to prove that $\lim\left((2n)^{1/n}\right) = 1$.
Here $n\in \mathbb{N}$. I have already proven that $(2n)^{1/n} > 1$
for $n > 1$. So we can write $(2n)^{1/n} = 1 + k$ for some $k > 0$ when
$n>1$. Hence $2n = (1+k)^n$ for $n>1$. By the Binomial theorem, if $n>1$, we have,
\[ 2n = 1+nk+ \frac{1}{2}n(n-1)k^2+\cdots + k^n \]
Because all terms in Binomial expansion are positive, we can write
\[ 2n \geqslant 1 + \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow \; 2n > \frac{1}{2}n(n-1)k^2 \]
\[ \Rightarrow 4 > (n-1) k^2 \]
\[ \Rightarrow k^2 < \frac{4}{n-1} \]
Above is true if $n>1$.
\[ \Rightarrow k^2 < \left(\frac{2}{\sqrt{n-1}}\right)^2 \cdots\cdots (1)\]
Now I use the following theorem I have already proven. Given $a\geqslant 0 $,
and $b\geqslant 0 $, then
\[ a<b \Longleftrightarrow a^2 < b^2 \Longleftrightarrow \sqrt{a}< \sqrt{b} \]
Now $n>1$, hence $(n-1) > 0$. Applying above theorem with $a=n-1$ and
$b=0$, we arrive at the conclusion that $\sqrt{n-1} > 0$. Hence $\frac{1}{\sqrt{n-1}} > 0$. Hence $\frac{2}{\sqrt{n-1}} > 0$.
Since $k>0$, again applying the above
stated theorem to the inequality $(1)$, we get
\[ k < \frac{2}{\sqrt{n-1}} \cdots\cdots (2)\]
for $n>1$. Now if we are given that $\varepsilon > 0$ is arbitrary, then $\frac{4}{\varepsilon^2}>0$.
\[1+\frac{4}{\varepsilon^2} > 1\]
By Archimedean Property, there exists a natural number $N_1$ such that
\[ 1< 1 + \frac{4}{\varepsilon^2} < N_1\]
\[\Rightarrow 0 < \frac{4}{\varepsilon^2} < N_1 - 1\]
\[ \Rightarrow \frac{\varepsilon^2}{4} > \frac{1}{N_1 - 1} > 0 \]
\[ \Rightarrow \left( \frac{\varepsilon}{2}\right)^2 > \left(\frac{1}{\sqrt{N_1 - 1}}\right)^2 \cdots\cdots(3)\]
Now $\frac{\varepsilon}{2} > 0$ and $N_1 >1$, so $N_1 - 1 > 0$. By above mentioned theorem, we have
$\sqrt{N_1 - 1} > 0$ , which implies $\frac{1}{\sqrt{N_1 - 1}}>0$. So applying the same theorem about the
square roots of nonnegative numbers to the equation $(3)$, we arrive at the conclusion that
\[\frac{\varepsilon}{2} > \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{N_1 - 1}} < \varepsilon \cdots\cdots(4)\]
Now for all $n \geqslant N_1$, because $N_1 > 1$, we have $n \geqslant N_1 > 1$.
\[ \Rightarrow n-1 \geqslant N_1 - 1 > 0 \]
\[ 0 < \frac{1}{n-1} \leqslant \frac{1}{N_1 - 1} \]
Again applying the same theorem, we get
\[ \frac{1}{\sqrt{n-1}} \leqslant \frac{1}{\sqrt{N_1 - 1}} \]
\[ \Rightarrow \frac{2}{\sqrt{n-1}} \leqslant \frac{2}{\sqrt{N_1 - 1}} \]
Combining with equation $(2)$ and equation $(4)$, we conclude that, for all $n\geqslant N_1$,
$k < \varepsilon$. But since $(2n)^{1/n} = 1 + k$ , and $k > 0$, we see that
\[ | (2n)^{1/n} - 1 | = k < \varepsilon \;\;\; \forall n \geqslant N_1 \]
Since $\varepsilon$ is arbitrary, this proves that $\lim\left((2n)^{1/n}\right) = 1$.Please let me know if proof sounds right.thanks

Perhaps you see why people detest "epsilon-delta" proofs so much, they are messy.

It looks correct, though I would make one small suggestion: your number $k$ depends on $n$ so I would notate it $k_n$ to indicate this dependence.

 

[issacnewton]

Deveno, this problem is from Bartle's Analysis book. He has not introduced limit definitions yet. At this point in the book, he has just finished basic definition of limit using epsilon-delta. So I am not supposed to use any limit theorems or L'Hospital's rule or log tricks. I am supposed to use epsilon-delta method. Proof certainly looks messy, but if I have to stick with epsilon-delta method, I can't make it more easy. The reason I am getting few responses is probably because of messiness of this proof.

 

[zzephod]

Prove that $$\lim_{n \to \infty} \ln( (2n)^{1/n} ) =0$$, you should only need L'Hopital's rule.

Except the limit is actually 1, not 0.

Read what was written.

.

 

[Deveno]

Deveno, this problem is from Bartle's Analysis book. He has not introduced limit definitions yet. At this point in the book, he has just finished basic definition of limit using epsilon-delta. So I am not supposed to use any limit theorems or L'Hospital's rule or log tricks. I am supposed to use epsilon-delta method. Proof certainly looks messy, but if I have to stick with epsilon-delta method, I can't make it more easy. The reason I am getting few responses is probably because of messiness of this proof.

Under those conditions, I don't see a problem. You have carefully justified the choice of your $N_1$, which is the crucial step. Furthermore, $N_1$ depends on $\epsilon$, which is to be expected.

As you may suspect, all proofs of this type involve "finding a bound that works". As long as the bound is *strict*, it doesn't matter what devious or convoluted means you use to obtain it (in other words many analytic proofs are "by hook or by crook").

Personally, I think you are to be commended on how diligently you put this together. These sorts of things are the "hardest part" of the calculus, and soon you shall have more sophisticated tools to make life much easier.

 

[issacnewton]

Deveno, thanks for the input. I think the reason inequalities are taught a lot in some pre-calculus courses is to prepare the student for these rigorous analysis proofs. I have
solved all problems from Daniel Velleman's How to prove it. It prepared me for the required rigor in analysis proofs.