Convergence of the series Ʃ(i^n)/n

[matt.qmar]

Hey,

I am trying to determine the convergence/divergence of

Ʃ_n=1^∞ iⁿ/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 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.

 

[jackmell]

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

 

[matt.qmar]

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]

Ʃ_n=1^∞ iⁿ/n = -log(1-i), woo hoo!

 

[jackmell]

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]

Ah, Thanks for the heads up. I suppose I sort of took that for granted. Would http://en.wikipedia.org/wiki/Abel's_test#Abel.27s_test_in_complex_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...

 

[jackmell]

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.