# Complex Power Series Convergence Help!

by two lock
Tags: complex analysis, convergence test, power series
 P: 5 1. The problem statement, all variables and given/known data 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. 2. Relevant equations As an example: Σ(z+3)k2 with k going from 0 → ∞ and z a complex number 3. 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!
 HW Helper Sci Advisor Thanks P: 24,455 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.
 P: 5 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: ∞ $\sum$(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.
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Thanks
P: 24,455

## Complex Power Series Convergence Help!

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?
 P: 5 Well when z=-2, we get: ∞ $\sum$1k2=1+1+...+1=∞ as k→∞ k=0 So ak does not tend toward zero. Then at z=-3+i: ∞ $\sum$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: ∞ $\sum$$\frac{z^{k}}{k^{2}}$ 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$_{k→∞}$ak=0 if the coefficients are convergent earlier in the problem set. Classic math professor, keeping things thorough.
 HW Helper Sci Advisor Thanks P: 24,455 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.
 P: 5 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: ($\frac{1}{n}$)$^{∞}_{n=1}$ doesn't converge? Or does it actually not?
HW Helper
 Quote by two lock 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: ($\frac{1}{n}$)$^{∞}_{n=1}$ doesn't converge? Or does it actually not?