Chapter 9 - Section 9.5 - Absolute Convergence; The Ratio and Root Tests - Exercises - Page 516: 53

Diverges

We are given that $a_{n+1}=\sqrt [n] a_n$ and $a_1=\dfrac{1}{3}$ Here, for $n=1,2,3,....n$ we have $a_2=\sqrt [1] a_1=(\dfrac{1}{3})^{1/1} \\ a_3=\sqrt [2] {\dfrac{1}{3}}=(\dfrac{1}{3})^{1/2}.... \\ ...a_{n+1}=(\dfrac{1}{3})^{1/n!} $ Thus, we have $l=\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{3})^{1/(n-1)!}=1$ So, $l=1$ Hence, the series Diverges with limit $1$.