Prove that $\lim ((2n)^\frac{1}{n}) = 1$.
I have obtained the following:
$$(2n)^\frac{1}{n} = 1 + k_{n}; n > 1$$ $$(2n) = (1 + k_{n})^n$$
By the Binomial Theorem: $$(1 + k_{n})^n = 1 + k_{n} + \frac{1}{2}n(n-1)k_{n}^2 + ...$$
So $$2n > \frac{1}{2}n(n-1)k_{n}^2$$ $$k_{n} < \frac{2}{\sqrt{n-1}}$$
However, I am not sure how to simplify this down to show $\frac{2}{\sqrt{n-1}} < \varepsilon, \forall \varepsilon > 0$ e.g. by invoking say $\frac{2}{\sqrt{n-1}} < \frac{1}{n} < \varepsilon$
$$1<(2n)^{\frac{1}{n}}<1+\frac{2}{\sqrt{n}}$$ works because by the binomial theorem we obtain: $$\left(1+\frac{2}{\sqrt{n}}\right)^n>1+2\sqrt{n}+\frac{4}{n}\cdot\frac{n(n-1)}{2}=2\sqrt{n}+2n-1>2n.$$ Now, $\frac{2}{\sqrt{n}}<\epsilon$ for $n>\frac{4}{\epsilon^2}.$
I will prove a nice theorem that is known as the Cauchy-D'Alembert criterion in my country(I have yet to find any references to it in English,so if you have any I would appreciate if you shared them with me).
Theorem : Let $(x_n)$ be a sequence of strictly positive real numbers such that $\exists \lim\limits_{n\to \infty} \frac{x_{n+1}}{x_n}$. Then we have $\lim \limits_{n\to \infty} \sqrt[n] x_n =\lim\limits_{n\to \infty} \frac{x_{n+1}}{x_n}$
Proof: Denote $L=\lim \limits_{n\to \infty} \sqrt[n] x_n$
$=>\ln L=\lim \limits_{n\to \infty}\frac{\ln x_n}{n}=\lim \limits_{n\to \infty}\frac{\ln x_{n+1}-\ln x_n}{n+1-n}=\lim \limits_{n\to \infty}\ln\frac{x_{n+1}}{x_n}$(I used the Stolz-Cesaro theorem).
Hence, $L=\lim \limits_{n\to \infty}\frac{x_{n+1}}{x_n}$.
Now, in your problem we have $x_n=2n$. By using the theorem above, we easily have that $\lim \limits_{n\to \infty}\sqrt[n] {2n}=\lim \limits_{n\to\infty}\frac{2n+2}{2n}=1$.
$$(2n)^{\frac{1}{n}}=e^{\frac{1}{n}\log(2n)}$$ Since ${\frac{1}{n}\log(2n)}\longrightarrow 0$, then $ {(2n)^{\frac{1}{n}}} \longrightarrow 1$
