# Radius of convergence of power series

1. Jan 14, 2007

### Benny

1. The problem statement, all variables and given/known data

Find the radius of convergence of the following series.

$$\sum\limits_{k = 1}^\infty {2^k z^{k!} }$$

2. Relevant equations

The answer is given as R = 1 and the suggested method is to use the Cauchy-Hadamard criterion; $$R = \frac{1}{L},L = \lim \sup \left\{ {\left| {a_k } \right|^{\frac{1}{k}} } \right\}$$

3. The attempt at a solution

I don't know where to begin. The sequence a_k in the Cauchy-Hadamard criterion is for series of the form $$\sum\limits_k^{} {a_k z^k }$$ but the series here has z raised to the power of k!, not just k. Substituting something for z (ie. set w = z^2 if the summation was over z^(2k)) doesn't work here. Can someone help me out? Thanks.

2. Jan 14, 2007

### matt grime

k! is a subsequence of n.

If it helps that series can be restated as

$$\sum_{n=1}^{\infty} a_nz^n$$

where a_n is given by:

if n=k!, then a_n=2^k
zero otherwise.

Clearly, terms that are zero have no effect on the limsup, so all you need to work out is

$$\lim\sup (a_n)^{1/n}$$

well, a_n is either zero, or if n=k!, then (a_n)^1/n = (2^k)^{1/k!}, so the lim sup is easy to work out (and is 1).

Last edited: Jan 14, 2007
3. Jan 14, 2007

### Benny

Thanks matt, I didn't think of looking at the series like that. Perhaps I could have started by considering the sequence of partial sums and the answer might have dropped out. Anyway, thanks again.