Radius on convergence in the complex plane

[racland]

Anyone knows anything about the Radius on convergence in the complex plane (Complex Analysis)

 

[DeadWolfe]

Yes. Somebody knows something about the radius of convergence in the complex plane.

Hope I was helpful

 

[ranger]

Yes. Somebody knows something about the radius of convergence in the complex plane.

Hope I was helpful

:rofl: :rofl:

 

[HallsofIvy]

Radius of convergence in the complex plane is exactly the same as radius of convergence in the real numbers. Exactly what is your question?

 

[racland]

Great

The problem states:
Find the Radius of Convergence of the following Power Series:
(a) Sumation as n goes from zero to infinity of Z^n!
(b) Sumation as N goes from zero to infinity of (n + 2^n)Z^n

For (a) I think the radius of convergence is 1 but I'm a bit unsure of that...

 

[HallsofIvy]

As I said- same thing as on the real line (except now it really is a radius!). Apply the "ratio" test: a series $\Sigma a_n$ converges absolutely if the limit ration
$$\lim_{n\rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$$
is less than 1.
(Diverges if that limit is greater than one, may converge absolutely or conditionally or diverge if it is equal to 1).

In particular, for Zⁿ, we have |Zⁿ⁺¹/Zⁿ|= |Z|. That series converges absolutely for |Z|< 1. (It obviously diverges for Z= 1 or -1 and diverges for |Z|> 1)

Try (n+ 2ⁿ)Zⁿ yourself.

 

[StatusX]

Is that $z^{n!}$? If so, you can prove a general result that if a_n is any increasing sequence of natural numbers, then $z^{a_n}$ converges iff |z|<1. This is the case HallsofIvy did if a_n=n, and yours if a_n=n!. The general proof follows from the result for a_n=n (which is the smallest increasing sequence of natural numbers).