Finding the Annulus of Convergence for a Laurent Series

[gestalt]

I am trying to understand the idea of annulus of convergence. This is the example I have been looking at but it has me completely stumped.

[∞]\sum[/n=1] (z^n!)(1-sin(1/2n))^(n+1)! + [∞]\sum[/n=1] (2n)!/[((n!)^2)(z^3n)]

All of the examples I have worked on in the past have been complex functions. This one seems odd because it is a Laurent Series.

 

[yus310]

I am trying to understand the idea of annulus of convergence. This is the example I have been looking at but it has me completely stumped.

[∞]\sum[/n=1] (z^n!)(1-sin(1/2n))^(n+1)! + [∞]\sum[/n=1] (2n)!/[((n!)^2)(z^3n)]

All of the examples I have worked on in the past have been complex functions. This one seems odd because it is a Laurent Series.

If have to go back to Calc II, and find or use a series of test series e.g.

alternating series
comparsion tests

to see how and when they converge...

 

[jackmell]

I am trying to understand the idea of annulus of convergence. This is the example I have been looking at but it has me completely stumped.

[∞]\sum[/n=1] (z^n!)(1-sin(1/2n))^(n+1)! + [∞]\sum[/n=1] (2n)!/[((n!)^2)(z^3n)]

All of the examples I have worked on in the past have been complex functions. This one seems odd because it is a Laurent Series.

I think your problem stems from evaluating the convergence of the term

\sum \frac{b_n}{z^n}

that has a region of convergence "greater" than some number. For example, suppose I let 1/z=w and consider:

\sum b_n w^n

and I can use any of the standard tests on that and find out it's radius of convergence is 3. That means

\left|\frac{1}{z}\right|<3

or:

|z|>1/3

That gives you the inner radius and the radius of convergence for the other sum gives you the outer radius.