If for all $n\geq N$ $$\sqrt[n]{a_n}<k<1$$ then $$a_n<k^n<1.$$ Since $\sum_{i=N}^\infty k^i$ converges so does $\sum_{i=N}^\infty a_n$ by the comparison test. If $\sqrt[n]{a_n}>1$ then by comparison with $\sum_{i=N}^\infty 1$ the series is divergent. Absolute convergence in case of nonpositive $a_n$ can be proven in exactly the same way using $\sqrt[n]{|a_n|}$ .