- #1

- 6

- 0

[∞]\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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter gestalt
- Start date

- #1

- 6

- 0

[∞]\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.

- #2

- 81

- 1

[∞]\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

- 1,800

- 53

[∞]\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.

Share: