ask Baby Rudin chapter 3 exercise 6(c)
Indicate the behavior (convergence or divergence) of $\Sigma a_n$ if
$$a_n = (\sqrt[n]{n} -1)^n$$
I solve this question by using the first solution. I don't understand the second solution. How to get $a_n \leq 2^{-n}$ for all large n?
Solution 1:
Since $\sqrt[n]{n}\rightarrow 1$, there exists $N$ such that $n>N$ implies $$0<\sqrt[n]{n}-1<\frac{1}{2}\mbox{,}$$ and hence $$a_n<2^{-n}\mbox{.}$$