Use the ratio test for the power series. With thus, Thus, the radius of convergence is the the reciprocal of that thus . However, the interval of convergence is . For this power series diverges.Originally Posted byTexasGirl

Now for the second problem, use the ratio test for power series again to get Thus, the radius of converges is 1, thus for converges absolutely. Checking the endpoint (because the ratio test is inconclusive for when its limit is one), we have

but this is the alternating-harmonic series thus it converges. For the second possibility we have that but this is the harmonic series thus it diverges. Thus, the interval of convergence for the second power series is