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

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

Hello

I 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

2. ## Re: Finding limit of (2n)^(1/n) where n is natural number

\displaystyle \begin{align*} \lim_{n \to \infty} \left[ \left( 2n \right) ^{\frac{1}{n}} \right] &= \lim_{n \to \infty} \left\{ e^{ \ln{ \left[ \left( 2n \right) ^{\frac{1}{n}} \right] } } \right\} \\ &= \lim_{n \to \infty} \left[ e^{\frac{1}{n} \ln{(2n)} } \right] \\ &= \lim_{n \to \infty} \left[ e^{ \frac{\ln{(2n)}}{n} } \right] \\ &= e^{ \lim_{n \to \infty} \left[ \frac{\ln{(2n)}}{n} \right] } \\ &= e^{\lim_{n \to \infty} \left( \frac{\frac{1}{n}}{1} \right) } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{n \to \infty} \left( \frac{1}{n} \right) } \\ &= e^0 \\ &= 1 \end{align*}

3. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Hello Proveit, thank for showing me that method, but I wanted to use basic definitions of limit as in analysis. So is my work Ok ? I am supposed to use only epsilon-delta method here....

4. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Originally Posted by issacnewton
, but I wanted to use basic definitions of limit as in analysis. So is my work Ok ? I am supposed to use only epsilon-delta method here....
That proof works even though it is very labored and hard to follow.

If for $n>2$ you let $s_n=\sqrt[n]{2n}-1$ then $2n=(s_n+1)^n$.

Now apply the binominal theorem: $2n\ge \dfrac{n(n-1)s_n^2}{2}\\ \frac{4}{n-1}\ge s_n^2$

It follows that $s_n\to 0$ or $\sqrt[n]{2n}\to 1$

5. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Thanks Plato....I think you are using Squeeze theorem here........The problem is from Bartle's analysis book and he has not yet introduced any limit theorems.... so in this section , student is supposed to use just epsilon-delta method......... maybe Bartle wants student to get the fill of all the labor in such proofs

6. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Originally Posted by issacnewton
I think you are using Squeeze theorem here........The problem is from Bartle's analysis book and he has not yet introduced any limit theorems.... so in this section , student is supposed to use just epsilon-delta method......... maybe Bartle wants student to get the fill of all the labor in such proofs
No, I am not really using that. It is clear that $|s_n-1|<c$ for any $c>0$ if we control $N$.
Now I have doubt that Bartle does do that, I do not have that textbook.
However, I have used his calculus text written for honors students.
Bartle is much like R L Moore, his method works wonderfully if one follows his program completely.
If it is as you report, then I suspect he is teaching students the strict $\epsilon /\delta$ method.

7. ## Re: Finding limit of (2n)^(1/n) where n is natural number

I don't know why you decided you had to truncate your binomial series to the square term. Remember, you are just trying to find a lower bound for n, so why not just say

\displaystyle \begin{align*} 2n &> 1 + n\,k \\ 2n - n\,k &> 1 \\ n \left( 2 - k \right) &> 1 \\ n &> \frac{1}{2 - k} \end{align*}

Then by letting \displaystyle \begin{align*} N = \frac{1}{2 - k} \end{align*} you should be able to show \displaystyle \begin{align*} n > N \implies \left| \left( 2n \right) ^{\frac{1}{n}} - 1 \right| < \epsilon \end{align*}, thus proving your limit.

Or even \displaystyle \begin{align*} 2n > 1 \implies n > \frac{1}{2} \end{align*} means you could set \displaystyle \begin{align*} N = \frac{1}{2} \end{align*}.

8. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Prove it, $\displaystyle N$ is a natural number, can't take fractional values.......But I get the point.......
Plato, I think Bartle expects the students to practice this $\displaystyle \varepsilon/\delta$ method, so as to learn to manipulate inequalities. The book is "Introduction to Real Analysis" by Bartle 4th edition.

9. ## Re: Finding limit of (2n)^(1/n) where n is natural number

There is no reason to believe that N should be a natural number. The \displaystyle \begin{align*} n > N \end{align*} part in showing limits to infinity really translates to "whatever big number you choose, your independent variable is going to be bigger than it."

10. ## Re: Finding limit of (2n)^(1/n) where n is natural number

Ok, that makes sense