Why is $1/n^{1/3}$ convergent? I thought because $p<1$ it would be divergent, but apparently not. Why is that?

You, certainly, know the series $\sum (1/n)$ and know that it is divergent. There is  a nice approach in which we can test the divergence or convergence. That is the Quotient Test or Limit comparison test. According to it, if $$\lim_{n\to\infty}\frac{u_n}{v_n}=A\neq0, ~~\text{or}~~ A=\infty$$ then $\sum u_n$ and $\sum v_n$ have the same destiny. Here, take $\sum v_n=\sum (1/n)$.
