Show that the sequence can be smaller than any $\varepsilon >0$

Question

How to find an upper-bound for the following sequence to show that it can be smaller than $\forall \varepsilon >0$: $$a_n=n^{1/n}-1$$

If we do the following operations: $$a_n = n^{1/n}-1 = e^{\ln n/n}-1>\frac{\ln n}{n}$$ we get unuseful information. I cannot find a workable upper-bound for the situation.

Any help is appreciated.

Answer 1

$a_n=e^{\frac 2 n \ln {\sqrt n}} -1<e^{\frac 2 n ({\sqrt n-1})} -1 <e^{\frac 2 n ({\sqrt n})} -1$. You can now find an explicit $N$ such that $a_n <\epsilon$ for $n >N$.

Answer 2

You don't need a special form for upper bound. Since $e^x-1\sim o(x)$, and $\lim\limits_{n\rightarrow \infty}\frac{\ln n}{n}=0$, the rest will be natural.