# Show that $\lim [(2n)^\frac{1}{n})] = 1$

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 $$\frac{2}{\sqrt{n-1}} < \frac{1}{n} < \varepsilon$$

• Whenever you fix $\varepsilon > 0$ there is $n$ such that $\frac{2}{\sqrt{n-1}} < \varepsilon$, just take the square of both sides and find $n$. So you should be done. – Gibbs Mar 23 at 19:56
• It would be nice if you let us know what is the variable in the expression and at which value of the variable the limit should be evaluated... – user Mar 23 at 21:27

$$1<(2n)^{\frac{1}{n}}<1+\frac{2}{\sqrt{n}}$$ 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 $$n>\frac{4}{\epsilon^2}.$$
$$(2n)^{\frac{1}{n}}=e^{\frac{1}{n}\log(2n)}$$ Since $${\frac{1}{n}\log(2n)}\longrightarrow 0$$, then $${(2n)^{\frac{1}{n}}} \longrightarrow 1$$
Hint: $$\lim (2n)^{\frac{1}{n}}=\lim e^{\frac{\ln(2n)}{n}}.$$
Theorem : Let $$(x_n)$$ be a sequence of strictly positive real numbers. 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 $$\lim \limits_{n\to \infty}\sqrt[n] {2n}=\lim \limits_{n\to\infty}\frac{2n+2}{2n}=1$$.