Complex Power Series Convergence Help

In summary, the original problem set asks for the radius of convergence of a complex series, and whether or not it converges anywhere on the radius of convergence. The Infinite Series Sum a_k Diverges if |a_k| does not approach 0. The new example provides an easier example in which the radius of convergence is at the origin and of radius 1. For any z such that z+3≥1, ak will not approach 0 as k increases towards infinity.
  • #1
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Homework Statement


I have a problem set that asks me to determine, first, the radius of convergence of a complex series (using the limit of the coefficients), and second, whether or not the series converges anywhere on the radius of convergence.

Homework Equations


As an example:
Σ(z+3)k2
with k going from 0 → ∞ and z a complex number

The Attempt at a Solution


I can figure out the radius of convergence easily enough (I think); it would be 1 here, right? My question is just about how to determine whether or not it converges on the circle of convergence. Honestly, I'm not even sure of how to test for convergence at a specific point.

My one thought was to plug in points on the circle, say z=-2 or z=-3+i in this case, but I'm not sure what the result would mean.

Thanks for any help you can provide!
 
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  • #2
An infinite series sum a_k diverges if |a_k| does not approach 0. What's |a_k| in your case of z being on the circle of convergence? BTW what it is the circle of converges. z=2 and z=3+i are not on it.
 
  • #3
Well that's partially why I used this example. I'm not sure how to handle when the exponent on the complex number is not just k. I realize now I made a mistake in the original post (I didn't make it on my problem set). If I'm right and the radius is 1, then what I meant to write was that z=-2 and z=-3+i are on the circle. Thanks.

Maybe a better (simpler) example is:

[itex]\sum[/itex](3k+1)zk
k=0

And to answer your question "what's |ak| in your case of z being on the circle of convergence?", I'm really not sure. In the new example I listed, the circle of convergence is at the origin and of radius 1, so points on it are just on the unit circle in the complex plane (z=1, i, -1, -i, etc.).

I guess if I had to take a shot in the dark, I would say that the size of the coefficients is unaffected whether or not it is on the unit circle.
 
  • #4
Stick with (3+z)^(k^2) for now. It's actually simpler than the other one. And, yes, now you've got two correct points on the circle of convergence. If you put z=(-2) can you give me a reason why the series diverges? Ditto for z=(-3+i)? The reason for them not converging is really the same. Can you generalize this to other points on the circle of convergence?
 
  • #5
Well when z=-2, we get:

[itex]\sum[/itex]1k2=1+1+...+1=∞ as k→∞
k=0
So ak does not tend toward zero.

Then at z=-3+i:

[itex]\sum[/itex]ik2
k=0
So again, ak does not tend toward zero.

In general, we can see that for any z such that z+3≥1, ak will not approach 0 as k increases towards infinity. That's a huge help! Thanks!

Just to make sure that I've got it, I would say the same thing for the other example. Unless |z|<1, the limit of the coefficients will approach ∞, -∞, or not exist, so the series converges nowhere on the circle of convergence. Whereas if we consider:

[itex]\sum[/itex][itex]\frac{z^{k}}{k^{2}}[/itex]
k=0
which has the same radius of convergence, we see that the series converges everywhere on the circle because the denominator of each consecutive term is going to be increasing while the numerator will not be, no matter what z is.

We actually proved that lim[itex]_{k→∞}[/itex]ak=0 if the coefficients are convergent earlier in the problem set. Classic math professor, keeping things thorough.
 
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  • #6
That's it. Except that just having the terms decreasing doesn't prove it converges. Think about z^k/k. It doesn't converge at z=1.
 
  • #7
I'm not sure why that wouldn't converge at z=1. As k increases infinitely, shouldn't 1/k become infinitely small?

Isn't saying that like saying that the sequence:
([itex]\frac{1}{n}[/itex])[itex]^{∞}_{n=1}[/itex]
doesn't converge? Or does it actually not?
 
  • #8
two lock said:
I'm not sure why that wouldn't converge at z=1. As k increases infinitely, shouldn't 1/k become infinitely small?

Isn't saying that like saying that the sequence:
([itex]\frac{1}{n}[/itex])[itex]^{∞}_{n=1}[/itex]
doesn't converge? Or does it actually not?

Check out http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
 

1. What is a complex power series?

A complex power series is a mathematical representation of a function, where the coefficients of the series are complex numbers. It is written as Σn=0 cn(z-z0)n, where cn is the coefficient and z is a complex variable.

2. How do you determine the convergence of a complex power series?

The convergence of a complex power series can be determined by using the ratio test or the root test. If the limit of the absolute value of the ratio or root of the series is less than 1, then the series is convergent. Otherwise, it is divergent.

3. What is the radius of convergence for a complex power series?

The radius of convergence for a complex power series is the distance from the center of the series (z0) to the nearest point where the series converges. It can be calculated using the ratio test or the root test.

4. Can a complex power series converge at more than one point?

Yes, a complex power series can converge at more than one point within its radius of convergence. However, it may also converge at all points on the boundary of the radius of convergence.

5. What is the significance of the radius of convergence in complex power series?

The radius of convergence determines the region of the complex plane where the series will converge. It indicates the set of complex numbers for which the series is a valid representation of the function. Outside of this region, the series is divergent.

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