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

In summary: 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
issacnewton
1,041
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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
 
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  • #2
IssacNewton said:
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 \(\displaystyle \lim_{n \to \infty} \ln( (2n)^{1/n} ) =0\), you should only need L'Hopital's rule.

.
 
  • #3
thanks for input zzephod. But is my present work correct ?
 
  • #4
zzephod said:
Prove that \(\displaystyle \lim_{n \to \infty} \ln( (2n)^{1/n} ) =0\), you should only need L'Hopital's rule.

.

Except the limit is actually 1, not 0.
 
  • #5
IssacNewton said:
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.
 
  • #6
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.
 
  • #7
Prove It said:
zzephod said:
Prove that \(\displaystyle \lim_{n \to \infty} \ln( (2n)^{1/n} ) =0\), you should only need L'Hopital's rule.


Except the limit is actually 1, not 0.

Read what was written.

.
 
  • #8
IssacNewton said:
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.
 
  • #9
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.

 

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

What is the formula for finding the limit of (2n)^(1/n) where n is a natural number?

The formula for finding the limit of (2n)^(1/n) is lim (2n)^(1/n) = 2. This means that as n approaches infinity, the value of (2n)^(1/n) will approach 2.

What is the significance of using natural numbers in this problem?

The use of natural numbers in this problem is significant because it allows us to explore the behavior of the expression (2n)^(1/n) as n becomes increasingly larger. Natural numbers, also known as counting numbers, are positive integers starting from 1. By using natural numbers, we can see how the expression approaches a specific value as n goes towards infinity.

How do you prove that the limit of (2n)^(1/n) is equal to 2?

To prove that the limit of (2n)^(1/n) is equal to 2, we can use the fact that the natural logarithm of 2 is equal to 0.693. By taking the natural logarithm of both sides of the equation, we get ln (lim (2n)^(1/n)) = ln 2. Then, using the property of logarithms, we can rewrite the left side as lim (1/n) ln (2n) = ln 2. As n approaches infinity, the value of ln (2n) also approaches infinity. This means that the left side of the equation becomes 0, proving that the limit is equal to 2.

Can this limit be extended to other expressions?

Yes, this limit can be extended to other expressions with similar forms, such as (an)^(1/n), where a is a constant. The value of the limit will still approach a as n goes towards infinity. However, it is important to note that this only works when the exponent is in the form of 1/n. Other exponents may result in different limits.

What real-world applications can this limit be used for?

This limit can be used in various real-world applications, such as calculating compound interest in finance, determining the growth rate of bacteria in biology, and analyzing the convergence of infinite series in mathematics. It can also be applied in computer science for generating fractal patterns and in physics for modeling the behavior of certain physical systems.

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