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Nested Sums

Date: 6/17/96 at 22:45:20
From: Anonymous
Subject: Nested sums
       infinity   1               infinity    1
Given    SUM     ---   = p   and    SUM      ---  = q
        j = 1    j^3               k = 1     k^2

Then evaluate 

 infinity   infinity     1
   SUM        SUM     --------      
  j = 1      k = 1    (j + k)^3

in terms of p and q.


Date: 6/18/96 at 6:36:28
From: Doctor Pete
Subject: Re: Nested sums

This was a really fun and interesting problem - thanks!  :)

I'll call your values p and q by their more common name, zeta(3) and 
zeta(2); these are particular values of the zeta function, which is 
defined as

               \---\  1
    zeta(z) =   >    ---  = Sum[1/k^z, {k,1,Infinity}].
               /---/ k^z

To simplify typing, I will use the latter notation for a sum.
(Incidentally, this is the way sums are expressed in Mathematica.)  
Then our problem is to find

  S = Sum[Sum[1/(k+j)^3, {k,1,Infinity}], {j,1,Infinity}]

in terms of zeta(2) and zeta(3).  Take the inside sum for a particular 
value of j (that is, think of j as being fixed).  Then we have

      Sum[1/(k+j)^3, {k,1,Infinity}]
    = - Sum[1/k^3, {k,1,j}] + Sum[1/k^3, {k,1,Infinity}]
    = - Sum[1/k^3, {k,1,j}] + zeta(3).


  S = Sum[-Sum[1/k^3, {k,1,j}]+zeta(3), {j,1,Infinity}].

Now, to evaluate this, we must think of the sum as a limit of a finite 
sum; that is,

      Sum[f(j), {j,1,Infinity}] = Lim[Sum[f(j), {j,1,N}], 


  S = Lim[Sum[-Sum[1/k^3, {k,1,j}]+zeta(3), {j,1,N}], N->Infinity]
    = Lim[N*zeta(3)-Sum[Sum[1/k^3, {k,1,j}], {j,1,N}], N->Infinity].

But note that this double sum can be re-expressed, since

      Sum[Sum[1/k^3, {k,1,j}], {j,1,N}]

    =   1/1^3
      + 1/1^3 +     1/2^3
      + 1/1^3 +     1/2^3 +     1/3^3
      + 1/1^3 +     1/2^3 +     1/3^3 + ... + 1/N^3
    =   N/1^3 + (N-1)/2^3 + (N-2)/3^3 + ... + 1/N^3

    = Sum[(N+1-j)/j^3, {j,1,N}]
    = Sum[(N+1)/j^3-1/j^2, {j,1,N}]
    = (N+1)Sum[1/j^3, {j,1,N}] + Sum[1/j^2, {j,1,N}].


  S = Lim[N*zeta(3)-(N+1)*Sum[1/j^3, {j,1,N}] + Sum[1/j^2, {j,1,N}]
    = -zeta(3)+zeta(2).

Another interesting problem is to show that zeta(2) = Pi^2/6.  There 
is a general formula for zeta(2n) for positive integer n in terms of 
Bernoulli numbers, but no similar expression has been found for odd 

Your above sum is approximately 0.442877165.

To be completely rigorous, however, one needs to show convergence 
under the transformations and rearrangements in the above solution; 
since this is probably not the intended purpose of the problem, I will 
not make these arguments.

If the notation I've described above is unclear, I might suggest you 
look at the following URL:


-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
College Calculus

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