The package EvaluateMultiSums -
some concrete examples

by Carsten Schneider (Carsten.Schneider@risc.jku.at, RISC, JKU Linz)

The packages EvaluateMultiSums requires the difference ring/field algorithms of the summation package Sigma. Thus we load it into Mathematica:

<<Sigma.m

before we enter the package EvaluateMultiSum:

<<EvaluateMultiSums.m;

Note 1: The package HarmonicSums is not needed if one does not deal with infinite summations. Thus we skip it in the beginning.

Note 2: In the following we present only some simple examples that can be carried out within minutes.
Large examples (up to 7-fold sums and summands with size up to 1 GB) mostly coming from my cooperation with DESY (Johannes Blümlein) can be found in the corresponding publications of my home page:
http://www.risc.jku.at/people/cschneid/publications.php

Multi-sums with finite summation ranges

Sum 1

[C. Krattenthaler, C. Schneider. Evaluation of binomial double sums involving absolute values. In: Algorithmic Combinatorics: Enumerative Combinatorics, Special Functions and Computer Algebra, in Honour of Peter Paule on his 60th Birthday, V. Pillwein, C. Schneider (ed.), Texts and Monographs in Symbolic Computuation, 2020. Springer, arXiv:1607.05314]

EvaluateMultiSum[Binomial[2 m,j+m]Binomial[2 n,i+n],{{i,0,j},{j,0,m}},{m,n},{0,0},{n,Infinity}]

Alternatively, one can also define first the sum:

mySum=SigmaSum[SigmaBinomial[2 m,j+m] SigmaSum[SigmaBinomial[2 n,i+n], {i,0,j}], {j,0,m}]

and can apply afterwards the command EvaluateMultiSum (this time with an empty summation range):

EvaluateMultiSum[mySum,{},{m,n},{0,0},{n,Infinity}]

Sum 2

[N. Henze and G. Last. Absent-minded passengers. Amer. Math. Monthly, 126(10):867--875, 2019. arXiv:1809.10192.
C. Schneider. The Absent-Minded Passengers Problem: A Motivating Challenge Solved by Computer Algebra. arXiv:2003.01921]

VV=((k (1-(1/n)))/n)+(Underoverscript[(∑^(.)), (i=1), (-k+n)])(((1-i-k+n) (1-((1-i-k+n)/(1-i+n))))/(1-i+n))+2 ((((-1+k) k)/(2 (-1+n) (n^2)))+(Underoverscript[(∑^(.)), (i=1), k])(Underoverscript[(∑^(.)), (j=1), (-k+n)])((((1-j-k+n)/(-j+n))-((1-j-k+n)/(1-j+n)))/n));

EvaluateMultiSum[VV,{},{k,n},{1,2},{n,Infinity}]

Sum 3

[C. Schneider, W. Zudilin. A case study for ζ (4).  arXiv : 2004.08158]

mySum=(Underoverscript[(∑^(.)), (j=1), n])-((2 (-(j^2) n-j (n^2)) ((((((({{n}, {j}})))^(.))))^2) (((({{j+2 n}, {n}})))^(.)) (Subscript[(1), -1+n]) (Subscript[(1), -j+n]) (Subscript[(2), -1+j+n]) (((-1))^(Underscript[j, .])) (((-1))^(Underscript[n, .])))/((j^3) n (((j+n))^2) (Subscript[(1+j), n]) (Subscript[(1+j), 2 n])));

EvaluateMultiSum[mySum,{},{n},{1},{Infinity}]

Sum 4

[J. Ablinger, J. Bluemlein, S. Klein, C. Schneider. Modern Summation Methods and the Computation of 2- and 3-loop Feynman Diagrams . Nucl. Phys. B (Proc. Suppl.) 205-206, pp. 110-115. 2010. arXiv::1006.4797]

summand= (-1)^r (-j+n-2)!Binomial[j+1,r] r!/(-j+n+r)!(-1)^s Binomial[-j+n+r-2,s]/((n-s) (s+1))

EvaluateMultiSum[summand,{{s,0,-j+n+r-2},{r,0,j+1},{j,0,n-2}},{n},{2}]

Sum 5

[C.Schneider, R.Sulzgruber. Asymptotic and exact results on the complexity of the Novelli - Pak - Stoyanovskii algorithm.Electron.J.Combin.24 (2), pp.1 - 33. 2017. arXiv : 1606.07597]

mySum=(Underoverscript[(∑^(.)), (j=1), (Subscript[λ, 2])])(Underoverscript[(∑^(.)), (k=0), (-1+j)])(((-1+j) (j-k) (((({{j+k}, {j}})))^(.)) (((({{-j-k+(Subscript[λ, 1])+(Subscript[λ, 2])}, {-j+(Subscript[λ, 1])}})))^(.)) S[1,-j-k+(Subscript[λ, 1])+(Subscript[λ, 2])])/(j+k));

The calculation takes around 8 minutes [stay tuned...]

EvaluateMultiSum[mySum,{},{(Subscript[λ, 2]),(Subscript[λ, 1])},{1,1},{(Subscript[λ, 1]),Infinity},SplitSums->False]

Multi-sums with infinite summation ranges

In the following we will use also the package

<<HarmonicSums.m

Sum 6

[C. Schneider, R. Pemantle. When is 0.999... equal to 1?. Amer. Math. Monthly 114(4), pp. 344-350. 2007.  arXiv:math/0511574]

summand=S[1,j](S[1,k+1]-1)/j/k/(k+1)/(k+j)

EvaluateMultiSum[summand,{{j,1,Infinity},{k,1,Infinity}}]

Here z2,z3,z5 are short cuts for the Riemann Zeta-values Zeta(2)=((π^2)/6), Zeta(3),Zeta(5), respectively.

Sum 7

[I. Bierenbaum, J. Bluemlein, S. Klein, C. Schneider .Two-Loop Massive Operator Matrix Elements for Unpolarized Heavy Flavor Production to O(epsilon). Nucl.Phys. B 803(1-2), pp. 1-41. 2008.  arXiv:0803.0273]

We introduce the Beta-function by

B[a_,b_]:=SigmaFactorial[a-1]SigmaFactorial[b-1]/SigmaFactorial[a+b-1]

and insert the summand

summand=B[N,k]/(k+N+2)S[1,k+N]^2

EvaluateMultiSum[summand,{{k,1,Infinity}},{N},{0},{Infinity}]

Calculation of  ep-expansions

Sum 8

[I. Bierenbaum, J. Bl ̈umlein, and S. Klein. Evaluating two-loop massive op-erator matrix elements with Mellin-Barnes integrals.Nucl. Phys. B (Proc.Suppl.), 160:85–90, 2006. 
P. Paule, B. Buchberger, L. Kartashova, M. Kauers, C. Schneider, F. Winkler. Algorithms in Symbolic Computation. In: Hagenberg Research, Bruno Buchberger et al. (ed.), Chapter 1, pp. 5-62. 2009. ]

We insert the summand

summand=Gamma[1+N] (((Gamma[1-ep] Gamma[ep] Gamma[1-2 ep+j] Gamma[1+ep+j] Gamma[1+k] Gamma[1+j+k+N])/(Gamma[1-ep+j] Gamma[2+j+k] Gamma[2+j+N] Gamma[2+k+N]))+((Gamma[-ep] Gamma[1+ep] Gamma[1-ep+j] Gamma[1+2 ep+j] Gamma[1+k] Gamma[1+ep+j+k+N])/(Gamma[1+j] Gamma[2+ep+j+k] Gamma[2+ep+j+N] Gamma[2+k+N])));

that depends on a small parameter ep and search for the first coefficients (Subscript[F, 0])[N],(Subscript[F, 1])[N],(Subscript[F, 2])[N] of its ep-expansion:

(Underoverscript[(∑^(.)), (k=1), ∞])(Underoverscript[(∑^(.)), (j=1), ∞])summand=(Subscript[F, 0])[N]+ep(Subscript[F, 1])[N]+(ep^2)(Subscript[F, 2])[N]+...

For concrete evaluations N=1,2,3,4 this can be accomplished with the function call

EvaluateMultiSum[summand,{{j,0,Infinity},{k,0,Infinity}},EvaluateIn->{N,1,4},ExpandIn->{ep,0,2}]

This means, e.g., that (Subscript[F, 1])[1]=-1,(Subscript[F, 1])[2]=-(1/2),(Subscript[F, 1])[3]=-(413/1296),(Subscript[F, 1])[4]=-(197/864).

Keeping N symbolic this can be accomplished with the function call (stay tuned, it takes about 8 minutes):

EvaluateMultiSum[summand,{{j,0,Infinity},{k,0,Infinity}},{N},{1},{Infinity},ExpandIn->{ep,0,2}]