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

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- Feb 29th 2008, 01:21 PMszpengchaofind its radius of convergence
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 - Feb 29th 2008, 06:08 PMmr fantastic
I think you'll find that $\displaystyle \lim_{n \rightarrow \infty} \frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}} = 1$.

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 ..... - Mar 1st 2008, 03:04 PMThePerfectHacker