Use the definition of convergent to prove :
$\displaystyle {(2n)}^{\frac {1}{n}} \to 1 $
$\displaystyle (2n)^{1/n} = 2^{1/n} n^{1/n} $. Obviously $\displaystyle 2^{1/n} \rightarrow 1 $, so we need to only prove that $\displaystyle n^{1/n} \rightarrow 1 $.
For n > 1 we have $\displaystyle n^{1/n} > 1^{1/n} = 1 $. So we can let $\displaystyle n^{1/n} = 1 + h_n $ where $\displaystyle h_n $ is some positive quantity depending on n.
So $\displaystyle n = (1 + h_n)^n $. Now, it is clear from the binomial theorem that $\displaystyle (1 + h_n)^n \ge 1 + nh_n + \frac{n(n+1)}{2} h_n \ge \frac{n(n+1)}{2} h_n $.
Can you show that $\displaystyle h_n $ goes to 0 and so $\displaystyle n^{1/n} = 1 + h_n $ goes to 1?