1) Let be be complex sequence. Say is such that is bounded. Let be an upper bound, so, for all . Let . Note, . But is a geometric series with so it converges. Theradius of convergenceis define to be ( or are included) so that if then we have convergence and if then we have divergence.

2)If then it means . Thus, if then . Pick to that . Then for sufficiently large we have that . Thus, but so the geometric series converges. By direct comparison test it follows that converges for . Similarly if then . Pick so that and so as in above and this is a divergent geometric series.

3)For (a) consider using #2 the radius of convergence is . If is any point on the circle then so it converges too. For (b) consider , it does not converge on the circle because if then , so by the divergence test it cannot converge. For (c) consider . At it converges but not at .