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Tricky Complex Series 
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#1
Feb1412, 01:57 PM

P: 8

Hey,
I am trying to determine the convergence/divergence of Ʃ_{n=1}^{∞} i^{n}/n. I have tried all the tests I could think of (Comparison, Ratio, Root, n^{th} term) and cannot determine it's convergence. If there was a formula, for say, the Mth partial sum S_{M} then, if the limit as M → ∞ of S_{M} is L, we have convergence to L but I can't seem to arrange for the thing to add up the first M terms. Clearly, I think something can be done with the fact that i^{n} = {i, 1, i, 1} repeatably with periodicity 4. I'm not sure how this can exactly be of help though! Any help appreciated, thanks. 


#2
Feb1412, 02:12 PM

P: 1,666

Do you know the power series for [itex]\log(1z)[/itex]? It has a raidus of convergence equal to one. However convergence on the boundary [itex]z=1[/itex] is well, what? Take a look at Wikipedia for "Radius of Convergence".



#3
Feb1412, 02:40 PM

P: 8

Ah! I see, and we have i = 1 but i ≠ 1 so there we go!
Thanks a bunch. Factoring out the 1 seems to make it all go quite nicely. 


#4
Feb1412, 02:41 PM

P: 8

Tricky Complex Series
Ʃ_{n=1}^{∞} i^{n}/n = log(1i), woo hoo!



#5
Feb1412, 03:01 PM

P: 1,666

You know, that's really not good enough. Why does it converge on the boundary except for z=1? Or rather prove that it does. If you wish anyway.



#6
Feb1412, 03:32 PM

P: 8

Ah, Thanks for the heads up. I suppose I sort of took that for granted. Would http://en.wikipedia.org/wiki/Abel%27...mplex_analysis works to show convergence for all z ≠ 1 on the boundary (in particular, at our point i ≠ 1), as the a_{n}'s are monotonically decreasing? Some the a_{n}, however, are negative...
Trying to take some z_{o} ≠ 1 with z_{o} = 1 and using some other test (root, comparison, etc.) doesn't seem to work either, though... 


#7
Feb1512, 05:27 AM

P: 1,666

Yeah, I don't know offhand how to show that. There are tests to check for convergence on the boundary. Would need to look into thoses. Don't thinik Abel test would apply though.



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