Laurent Series Convergence

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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.
 

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  • #2
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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...
 
  • #3
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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

[tex]\sum \frac{b_n}{z^n}[/tex]

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

[tex]\sum b_n w^n[/tex]

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

[tex]\left|\frac{1}{z}\right|<3[/tex]

or:

[tex]|z|>1/3[/tex]

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

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