Re: idea 1 for Besançon atelier

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Fri, Oct 06, 2017 at 09:59:00PM +0200, Vincent Delecroix wrote:
> I tried with more or less success to get numerical approximations in PARI/GP
> (sumnummonien is sometimes very wrong and sumnum is sometimes very slow).
> Due to the very specific form of my numbers I believe that there are more
> clever algorithms.
> 
> In case I succeded to get approximations, I just use the very nice lindep
> function to check relations with standard multizetas or product of zeta.

It seems to me you can try to express your sum as linear combinaison of 
multi-zeta by rewriting the rational functions as linear combination
of rational function of a special form.
I give an example

Let [a,b,c]=sum 1/(n^a*(n+m)^b*m^c)
(which converges if a+b>=2 and b+c>=2)
Then
[a,b,c] = [c,b,a]
[a,0,c] = zeta(a)*zeta(c)
[a,b,0] = zetamult([b,a])

By summing the trivial identity
1/(m^a*(n+m)^b*n^(c-1)) + 1/(m^(a-1)*(n+m)^b*n^c) = 1/(m^a*(n+m)^(b-1)*n^c)
we have 
[a,b,c-1]+[a-1,b,c]=[a,b-1,c]
By applying this formula we get all the triples with a+b+c=6

[1,5,0]+[0,5,1] = [1,4,1] so [1,4,1] = 2*zetamult([5,1])
[1,4,1]+[0,4,2] = [1,3,2] so [1,3,2] = 2*zetamult([5,1])+zetamult([4,2])
[1,3,2]+[0,3,3] = [1,2,3] so [1,2,3] = 2*zetamult([5,1])+zetamult([4,2])+zetamult([3,3])
[1,2,3]+[0,2,4] = [1,1,4] so [1,1,4] = [1,2,3]+zetamult([2,4])
[1,3,2]+[2,3,1] = [2,2,2] so [2,2,2] = 4*zetamult([5,1])+2*zetamult([4,2])
[1,2,3]+[2,2,2] = [2,1,3] so [2,1,3] = [1,2,3]+[2,2,2]

By continuing one last step, we get the two missing ones:

[2,1,3]+[3,1,2] = [3,0,3] so zeta(3)^2 = something in term of zetamult

[1,1,4]+[0,1,5] = [1,0,5]
where both [0,1,5] and [1,0,5] are diverging multizeta value.
It is remarkable this sum can be interpreted as the difference of
two diverging multizeta values.

Cheers,
Bill.