Could someone please help me out with the following? I need to determine the radius of convergence of the following series. It is exactly as given in the question.

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

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.

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HallsofIvy
Homework Helper
Are you saying that a 'constant' sequence, like 3 , 3, 3, 3, ..., does not converge? Hmm, given $\epsilon> 0$, how would I find "N" so that if n> N, $|a_x- 3|< \epsilon$? Looks to me like, since |3- 3|= 0< $\epsilon$, any N would work. 3 is not a "limit point" in the topological sense but it definitely is a limit of that sequence. However, 1 and 3 are NOT subsequential limits of the sequence $(3+ (-1)^n)^n$. When n is odd, we have (3+ (-1)n)= 3- 1= 2 and when n is even (3+ (-1)n)= 3+ 1= 4. The two subsequential limits are 2 and 4.

Hmm...I mixed up questions from two sources because they looked similar. The coefficient should've had a 2 in place of the 3, my bad.

(2+(-1)^n) alternates between 1 and 3 but I don't see how the Cauchy-Hadamard criterion can be applied here. I suppose that for sequences with a finite number of different terms I only need to work out limits of the subsequences.

AKG
Homework Helper
I think you're over complicating this.

$$\limsup _{n \to \infty} (2 + (-1)^n) = 3$$

HallsofIvy