## Thomas' Calculus 13th Edition

As we are given that $a_{n+1}=\sqrt [n] a_n$ and $a_1=\dfrac{1}{3}$ ; and $n=1,2,3,....n$ Now, we can see that $a_2=\sqrt [1] a_1=(\dfrac{1}{3})^{1/1} \\ a_3=\sqrt [2] {\dfrac{1}{3}}=(\dfrac{1}{3})^{1/2}.... \\ ...a_n=(\dfrac{1}{3})^{(1/n)!}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{3})^{(1/n!)}=1$ Thus, the series diverges with limit $1$.