Re: Evaluating an infinite series

From: Julian V. Noble (jvn_at_virginia.edu)
Date: 06/22/04 

Date: Tue, 22 Jun 2004 14:16:32 -0400

"Julian V. Noble" wrote:
>
> "Brian R." wrote:
> >
> > Hello, all -- I need a little help with this problem, using Residues
> > to find this sum:
> >
> > Inf.
> > Sum 1/[(3k-2)^2] = ?
> > k=1
> >
> > Any help would be appreciated!
> >
> > Thanks.
>
> To evaluate Sum{k=0 to \infty}{f(k)} by residues, look at the
> contour integral
>
> \oint{ dz f(z) cot(z) }
>
> on the circle of radius R = N + 1/2 .
>
> You see that this will give you (up to trivial factors)
> twice the sum you want, plus the residue at z=0 and at any
> poles of f(z). This, of course, has to equal the actual integral
> on the contour.
>
> I leave it to you as a HW problem to show that the integral
> on the circle shown (which goes through none of the poles)
> vanishes in the limit N \to \infty .
>
> BTW, this is all found in standard books on functions of a complex
> variable, such as Copson or Whittaker & Watson, as well as in many
> books on mathematical methods of physics. I am thinking in particular
> of Riley, et al.
>

Oops-- two corrections:

        1. Integrate \oint{ dz f(z) cot(z \pi) } around |z| = N + 1/2
                                          ^^^

        2. The result will give you the sums

                \sum{k=1 to N} \left( { f(k) + f(-k) } \right)

           one of which is the sum you want and the other is one you have to
           evaluate by a trick.

But anyway, you get the general idea. 

-- 
Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
    ^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
   "For there was never yet philosopher that could endure the toothache
      patiently."  -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1.

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