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Tricky Complex Series

by matt.qmar
Tags: complex, series, tricky
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matt.qmar
#1
Feb14-12, 01:57 PM
P: 8
Hey,

I am trying to determine the convergence/divergence of

Ʃn=1 in/n.

I have tried all the tests I could think of (Comparison, Ratio, Root, nth term) and cannot determine it's convergence.

If there was a formula, for say, the Mth partial sum SM then, if the limit as M → ∞ of SM 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 in = {i, -1, -i, 1} repeatably with periodicity 4. I'm not sure how this can exactly be of help though!

Any help appreciated, thanks.
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jackmell
#2
Feb14-12, 02:12 PM
P: 1,666
Do you know the power series for [itex]\log(1-z)[/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".
matt.qmar
#3
Feb14-12, 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.

matt.qmar
#4
Feb14-12, 02:41 PM
P: 8
Tricky Complex Series

Ʃn=1 in/n = -log(1-i), woo hoo!
jackmell
#5
Feb14-12, 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.
matt.qmar
#6
Feb14-12, 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 an's are monotonically decreasing? Some the an, however, are negative...

Trying to take some zo ≠ 1 with |zo| = 1 and using some other test (root, comparison, etc.) doesn't seem to work either, though...
jackmell
#7
Feb15-12, 05:27 AM
P: 1,666
Yeah, I don't know off-hand 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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