Nested SumsDate: 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. Thanks, -Michael 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 oo \---\ 1 zeta(z) = > --- = Sum[1/k^z, {k,1,Infinity}]. /---/ k^z k=1 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). So 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}], N->Infinity]. Therefore 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}]. Hence S = Lim[N*zeta(3)-(N+1)*Sum[1/j^3, {j,1,N}] + Sum[1/j^2, {j,1,N}] N->Infinity] = -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 values. 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: http://www.ugcs.caltech.edu/~peterw/studies/bernoulli/ -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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