- #1

- 584

- 0

[tex]

\sum\limits_{n = 0}^\infty {\left( {3 + \left( { - 1} \right)^n } \right)^n } z^n

[/tex]

The suggestion is to use the Cauchy-Hadamard criterion. The nth coefficient of this series is a_n = (3+(-1)^n)^n which is positive so |a_n|^(1/n) = (3 + (-1)^n). At first thought there are two limit points of the set of points of |a_n|^(1/n), 1 and 3. So the radius of convergence is R = 1/(limpsup(...)) = (1/3) which is the answer that is given.

The problem is that the set of points of the sequence (3 + (-1)^n) only consists of two points, 1 and 3. So how can it have any limit points? (No neighbourhood of either of these two points contains an 'infinite' number of points of the set since there are only two different points.)

Can someone please explain how to do this question properly? Thanks.