# find its radius of convergence

• February 29th 2008, 01:21 PM
szpengchao
sigma ( n^(sqrt(n)) * z^n, n>= 1, z is complex)

how to find its radius of convergence, it tried the ratio test
but it doesnt work
• February 29th 2008, 06:08 PM
mr fantastic
Quote:

Originally Posted by szpengchao
sigma ( n^(sqrt(n)) * z^n, n>= 1, z is complex)

how to find its radius of convergence, it tried the ratio test
but it doesnt work

I think you'll find that $\lim_{n \rightarrow \infty} \frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}} = 1$.

So the ratio test forces the requirement |z| < 1. Then you must test the cases z = 1 and z = -1 for convergence, which should be very simple to do .....
• March 1st 2008, 03:04 PM
ThePerfectHacker
Quote:

Originally Posted by mr fantastic
So the ratio test forces the requirement |z| < 1. Then you must test the cases z = 1 and z = -1 for convergence, which should be very simple to do .....

No. $z$ is complex so $|z|=1\implies z = e^{i\theta}$. There are infinitely many points to check.