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#### issacnewton

##### Member

- Jan 30, 2012

- 61

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