Difficult radias of convergence problem

[jessicaw]

Homework Statement

\sum_{m=0}^{\infty}i^{m}z^{m!}

Homework Equations

cauchy hadamard formula and limsup

The Attempt at a Solution

as the power of z is m! it becomes extremely hard.
i try to divide it into subseuqnce where "n=m!" and use the limsup formula but i get confused.
Should i use 1/m or 1/m! in calculating limsup?

sorry hallsofivy it is edited now

 

[HallsofIvy]

If that is really what you have- that is the sum is over n but only m occurs in the formula: \sum_{n=0}^{\infty}i^{m}z^{m!}
then it the sum of a constant, i^mz^m+ i^mz^m+ \cdot\cdot\cdot and so only converges for z= 0. Its radius of convergence is 0.

But your sum is probably intended to be \sum_{n=0}^{\infty}i^{n}z^{n!}

By the "ratio test", that will converges as long as
\frac{|i^{n+1}z^{n+1}|}{|i^n z^{n}|}= |z|< 1
and so its radius of convergence is 1.

 

[micromass]

So Cauchy-Hadamard tells us that we need to calculate \limsup_{n\rightarrow+\infty}{\sqrt[n]{c_n}}

In this case we have that the sequence c_n is

0,i^0+i^1,i^2,0,0,0,i^3,0,0,...,0,0,i^m,0,0,...

So we want the limsup of

0,\sqrt[1]{|i^0+i^1|},\sqrt[2]{|i^2|},0,0,0,\sqrt[6]{|i^3|},0,0,...,0,0,\sqrt[m!]{|i^m|},0,0,...

The zeros can surely be left out as they will not affect the limsup. So you would want to calculate

\limsup_{n\rightarrow+\infty}{\sqrt[n!]{|i^n|}. Which seems to be 1 in my opinion.