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Proving limit of a given sequence

issacnewton

Member
Jan 30, 2012
61
Hello


I want to prove the following.
\[ \lim_{n\rightarrow \infty}\left((2n)^{1/n}\right) = 1 \]
where \( n \in \mathbb{N} \). Now since we have nth root of a positive number,
I used theorem on the existence of nth root to argue that \( (2n)^{1/n} > 0 \).
Next I tried to prove that \( (2n)^{1/n} > 1 \) as following. Assume that
\( (2n)^{1/n} \leqslant 1 \). If \( (2n)^{1/n} = 1 \) then \( 2n = 1\Rightarrow \; n = \frac{1}{2} \) , which is a contradiction since


\( n\in \mathbb{N} \). So if \( (2n)^{1/n} < 1 \) then \( 0< (2n)^{1/n} < 1 \).
Then we have
\[ (2n)^{1/n} = \frac{1}{1+a} \]
for some \( a > 0 \). It follows that
\[ 2n = \frac{1}{(1+a)^n} \]
\[ \Rightarrow \; 2n = \frac{1}{1+na+\ldots} \]
Since \( a>0 \), we have \( (1+na + \ldots ) > 1 \). So
\[ \frac{1}{1+na+\ldots} < 1 \]
\[ \Rightarrow \; 2n < 1 \]
which means \( n < \frac{1}{2} \). Since \( n\in \mathbb{N} \), this
leads to contradiction. So I proved that \( (2n)^{1/n} > 1 \). If this is so,
then it can be written as
\[ (2n)^{1/n} = 1 + k \]
for some \( k > 0 \).
\[ \Rightarrow \; 2n = (1+k)^n \]
Till this far I have come. I am thinking of either using Binomial theorem or
Bernoulli's inequality. Any guidance will be helpful.


Thanks
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello


I want to prove the following.
\[ \lim_{n\rightarrow \infty}\left((2n)^{1/n}\right) = 1 \]
where \( n \in \mathbb{N} \). Now since we have nth root of a positive number,
I used theorem on the existence of nth root to argue that \( (2n)^{1/n} > 0 \).
Next I tried to prove that \( (2n)^{1/n} > 1 \) as following. Assume that
\( (2n)^{1/n} \leqslant 1 \). If \( (2n)^{1/n} = 1 \) then \( 2n = 1\Rightarrow \; n = \frac{1}{2} \) , which is a contradiction since


\( n\in \mathbb{N} \). So if \( (2n)^{1/n} < 1 \) then \( 0< (2n)^{1/n} < 1 \).
Then we have
\[ (2n)^{1/n} = \frac{1}{1+a} \]
for some \( a > 0 \). It follows that
\[ 2n = \frac{1}{(1+a)^n} \]
\[ \Rightarrow \; 2n = \frac{1}{1+na+\ldots} \]
Since \( a>0 \), we have \( (1+na + \ldots ) > 1 \). So
\[ \frac{1}{1+na+\ldots} < 1 \]
\[ \Rightarrow \; 2n < 1 \]
which means \( n < \frac{1}{2} \). Since \( n\in \mathbb{N} \), this
leads to contradiction. So I proved that \( (2n)^{1/n} > 1 \). If this is so,
then it can be written as
\[ (2n)^{1/n} = 1 + k \]
for some \( k > 0 \).
\[ \Rightarrow \; 2n = (1+k)^n \]
Till this far I have come. I am thinking of either using Binomial theorem or
Bernoulli's inequality. Any guidance will be helpful.


Thanks
May be that Your task is simplified if You try to find...

$\displaystyle \lim_{n \rightarrow \infty} \ln\ \{ (2\ n)^{\frac{1}{n}} \}\ (1)$

Kind regards

$\chi$ $\sigma$
 

issacnewton

Member
Jan 30, 2012
61
Hello

The book I am doing these problems from (Bartle 4ed.), has not introduced any complicated theorems about limits so far (This is section 3.1 in the chapter on limits, which is 3rd chapter). So I am supposed to use some basic machinery developed so far, which is definition of limit, Bernoulli's inequality, Binomial theorem, Archimedes's theorem, Completeness property etc. Using logarithms would be outside of this scope as the author has not used it till this point. I am sure there is some way of handling this within the basic machinery.