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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.

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/
```
Associated Topics:
College Calculus

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