Investigate the behavior of the sequence $a_n=(n^{\frac{1}{n}}-1)^{n}$

Question

My attempt: Let $n=(1+x_n)^{n}$ then $n\ge (1+\frac{n(n-1)}{2}x_n)$

$x_n \le \frac{ (2)}{n}$

$0<(x_n)^n<( \frac{2}{n})^n$

I am stuck after this .. can someone help me from this step onwards

Answer 1

We know from here that

$$\lim_{n\to\infty}n^{1/n}=1$$

This implies there exists $N$ such that $n\geq N$ implies

$$n^{1/n}<\frac{3}{2}$$

Then for $n\geq N$ we have

$$a_n=(n^{1/n}-1)^n<\left(\frac{3}{2}-1\right)^n=\frac{1}{2^n}$$

Since $n^{1/n}\geq 1$, this implies

$$0\leq (n^{1/n}-1)^n=a_n<\frac{1}{2^n}$$

for all $n\geq N$. By the squeeze theorem, we conclude

$$\lim_{n\to\infty}a_n=0$$