If for all $n\geq N$

then

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 $\sqrt[n]{|a_{n}|}$.