Divergence of Complex Harmonic

  • #1

Homework Statement



WTS divergence of:

[tex]\sum_{-\infty}^{\infty} \frac{1}{z-n}[/tex]



Homework Equations



Basic algebraic manipulation, standard tests of non-convergence?

The Attempt at a Solution



I have been playing with algebra, perhaps this equivilant (I hope equivilant, anyway) series may be of use:

[tex]\frac{1}{z}+\sum_{1}^{\infty} \frac{1}{z-n}+\frac{1}{z+n} = \frac{1}{z}+\sum_{1}^{\infty} \frac{2z}{z^2-n^2}[/tex]

Is this right? If so, what can I do from here? If not, where am I going wrong?

Thanks!
 

Answers and Replies

  • #2
Actually, now I'm thinking it may converge, unless I made a mistake.

[tex]\left | \sum \frac{2z}{z^2-n^2} \right | \le \sum \left | \frac{2z}{z^2-n^2} \right |[/tex]

Let [tex]z=re^{i\theta}[/tex], then


[tex]\left | \frac{2z}{z^2-n^2} \right | \le \frac{2r}{n^2-r^2}[/tex].

Since the real series

[tex]\sum \frac{2r}{n^2-r^2}[/tex]

converges, we have that the complex series converges absolutely for all z not a real integer.

Does this seem plausible, or did I trip up somewhere?
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
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6
In order to claim that

[tex]\sum_{n=-\infty}^{\infty}\frac{1}{z-n}=\frac{1}{z}+\sum_{n=1}^{\infty} \frac{1}{z-n}+\frac{1}{z+n} = \frac{1}{z}+\sum_{n=1}^{\infty} \frac{2z}{z^2-n^2}[/tex]

you first need to know that the original sum converges. If it doesn't converge (and it doesn't), you can't re-order terms like this.

Instead, try the most basic test for divergence of an infinite sum; the limit test.
 
  • #4
IThanks for the reply. I'm working out of a pretty difficult (IMO) book. It has no sort of standard calculus limit tests... not that I know how to apply it to this kind of problem anyway (with the dual infinity bounds for n).

I was thinking maybe I could make use of this: The book derives the following

[tex]\pi \cot \pi z = \frac{1}{z} - \sum_{n = 1}^{\infty} \left [ \frac{1}{n-z} - \frac{1}{n+z}\right ] = \frac{1}{z}- \sum_{n = 1}^{\infty} \frac{2z}{n^2-z^2}[/tex]

So if I assume BWOC that it does converge, I may be to combine it with the above somehow and obtain a contradiction? Not sure really.
 
  • #5
I don't see why I can't do the reordering:

[tex]\lim_{n\to \infty} \sum_{-N}^N \frac{1}{z-n} = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \sum_{-N}^{-1} \frac{1}{z-n} \right)[/tex]

[tex] = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \sum_{1}^N \frac{1}{z+n} \right )[/tex]

[tex] = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \frac{1}{z+n} \right )[/tex]
 
  • #6
1,838
7
When you group together +n an -n then the sum converges to the pi cot(pi z) function. That's quite easy to prove as the poles and the residues at the poles coincide. The difference must then be an analytic function but it is easy to see that this function is bounded, hence it must be a constant (in this case equal to zero) by Liouville's theorem.
 
  • #7
gabbagabbahey
Homework Helper
Gold Member
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In general,

[tex]\sum_{n=-\infty}^{\infty}a_n\neq \lim_{N\to \infty} \sum_{n=-N}^N a_n[/tex]

When you do this, are essentially reordering the terms in the sum so they are symmetric about [itex]n=0[/itex], and you can't reorder terms until you prove that the sum converges.

What convergence/divergence tests are discussed in your text (which text are you using BTW?)?
 
  • #8
1,838
7
I don't see why I can't do the reordering:

[tex]\lim_{n\to \infty} \sum_{-N}^N \frac{1}{z-n} = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \sum_{-N}^{-1} \frac{1}{z-n} \right)[/tex]

[tex] = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \sum_{1}^N \frac{1}{z+n} \right )[/tex]

[tex] = \lim_{n\to \infty} \frac{1}{z} + \left (\sum_{1}^N \frac{1}{z-n} + \frac{1}{z+n} \right )[/tex]

You are computing the limit in a special way here. If you make the lower limit -M and the upper limit N and take the two independent limits of M and N going to infinity, then the limit does not exist.
 
  • #9
In general,

[tex]\sum_{n=-\infty}^{\infty}a_n\neq \lim_{N\to \infty} \sum_{n=-N}^N a_n[/tex]

When you do this, are essentially reordering the terms in the sum so they are symmetric about [itex]n=0[/itex], and you can't reorder terms until you prove that the sum converges.

What convergence/divergence tests are discussed in your text (which text are you using BTW?)?

Ok... so my "intuition" to split them at zero symmetrically is where I went wrong. I am using Complex Analysis: In the Spirit of Litman Bers. This question must be related to 7.3 "the cotangent function." No tests of any kind are talked about in detail in this chapter. Chapter 7 is about metrics and sequences of holomorphic functions.

You are computing the limit in a special way here. If you make the lower limit -M and the upper limit N and take the two independent limits of M and N going to infinity, then the limit does not exist.

I see, so is this the right route for me to take then? I have never attempted two independent yet simultaneous limits on a sum's bounds before... the concept is nothing I've seen before.
 
  • #10
1,838
7
It depends on what you want to do. If you want to represent the Cotangent function in terms of the Mittag Leffler expansion, you have to write down the symmentrical summation as you did.
 
  • #12
1,838
7
The series as you wrote it down converges to pi cot(pi z). There is no contradiction with the fact that it is similar to the harmonic series (which we all know does diverge), because you take the upper and lower limits symmetrically.

Compare e.g. with the integral from minus to plus infinity of x dx

This diverges, because the limit of R1 and R2 to infinity of the integral from minus R1 to plus R2 diverges. But the symmetric integral from minus R to plus R is zero, so the limit of R to infinity of this symmtric integral does exist and is equal to zero.
 

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