Compute $\lim _{n \rightarrow \infty}(\sqrt[n]{n}-1)^{\frac{1}{n}}$.

I need to solve the following problem: $$ \lim _{n \mapsto \infty}(\sqrt[n]{n}-1)^{\frac{1}{n}}. $$

My attempt: $$\lim _{n \mapsto \infty}\ln(\sqrt[n]{n}-1)^{\frac{1}{n}}=\lim _{n \mapsto \infty}\frac{\ln(\sqrt[n]{n}-1)}{n}.$$ Now I want to prove that $$\lim _{n \mapsto \infty}\frac{\ln(\sqrt[n]{n}-1)}{n}=0.$$ But I'm stuck on proving the limit without the Heine theorem (equivalence relation between limit of a sequence and limit of a function). Any help would be greatly appreciated.

Solution 1:

Let $n$ be a positive integer.

Using Landau notations, $\sqrt[n]{n} -1= \exp(\frac{1}{n}\ln(n)) -1= \frac{\ln(n)}{n}+ o\left(\frac{\ln(n)}{n}\right)$ as $n\to+\infty$

Then $\frac{1}{n}\ln(\sqrt[n]{n}-1) = \frac{1}{n}\ln\left(\frac{\ln(n)}{n}+o\left(\frac{\ln(n)}{n}\right)\right) \to 0$ when $n\to+\infty$

And $(\sqrt[n]{n}-1)^\frac{1}{n}=\exp(\frac{1}{n}\ln(\sqrt[n]{n}-1)) \to \exp(0)=1$ when $n\to+\infty$

Solution 2:

We have for n>2

$\frac{\ln(\sqrt[n]{n}-1)}{n} = \frac{\ln\left[\sqrt[n]{n}\left(1-\frac{1}{\sqrt[n]{n}}\right)\right]}{n}$

$= \frac{\ln(\sqrt[n]{n})}{n}+\frac{\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)}{n}$

$= \frac{\ln(n)}{n^2}+\frac{\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)}{n}$

But for x>1, $\ln(x)\leq x-1$ so

$0\leq \frac{\ln(n)}{n^2} \leq \frac{n-1}{n^2}$

and the sequence $\left(\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)\right)_{n\geq 2}$ is bounded...