Double Series

Author

Eric W. Weisstein
January 17, 2009

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Algebra/DoubleSeries.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/DoubleSeries.html.

©2009 Wolfram Research, Inc. except for portions noted otherwise

Additional Authors

Examples from Antov Antonov
Dec 31, 2001

Code

DoubleSeries[s_,{m_,n_}]:=Sum[s,{i,∞},{j,∞}]+Sum[s,{i,∞},{j,-1,-∞,-1}]+Sum[s,{i,-1,-∞,-1},{j,1,∞}]+Sum[s,{i,-1,-∞,-1},{j,-1,-∞,-1}]+Sum[s/.i->0,{j,-∞,-1}]+Sum[s/.i->0,{j,∞}]+Sum[s/.j->0,{i,∞}]+Sum[s/.j->0,{i,∞}]

NDoubleSeries[s_,{m_,n_}]:=NSum[s,{i,∞},{j,∞}]+NSum[s,{i,∞},{j,-1,-∞,-1}]+NSum[s,{i,-1,-∞,-1},{j,1,∞}]+NSum[s,{i,-1,-∞,-1},{j,-1,-∞,-1}]+NSum[s/.i->0,{j,-∞,-1}]+NSum[s/.i->0,{j,∞}]+NSum[s/.j->0,{i,∞}]+NSum[s/.j->0,{i,∞}]

Writing a Double Series as a single series

(∑_(i=1)^m)(∑_(j=1)^n)x(i) y(j)=((∑_(i=1)^m)x(i))( (∑_(i=1)^n)y(i))

DoubleSeries[m_,n_]:=Module[{
s1=Plus@@Flatten[Table[x[i]y[j],{i,m},{j,n}]],s2=Plus@@Table[x[i],{i,m}]Plus@@Table[y[i],{i,n}]
},
{s1,s2,Simplify[s1-s2]}
]

DoubleSeries[1,3]//ColumnForm

DoubleSeries[2,3]//ColumnForm

DoubleSeries[3,3]//ColumnForm

Writing a Double Series as a single series

(∑_(l=0)^∞)(∑_(k=0)^∞)(Subscript[c, k,l])=(∑_(i=0)^∞)(Subscript[b, i])

b[n_]:=Array[c,{1,n+1},{n,0},Plus]+Array[c,{n,1},{0,n},Plus]

Table[b[n],{n,0,3}]/.c[x__]:>Subscript[c,x]//ColumnForm//TraditionalForm

c3=Plus@@Flatten[Table[c[i,j],{i,0,3},{j,0,3}]]

b3=Sum[b[i],{i,3}]

c3==b3

Lattice Sums

DoubleSeries[((((-1))^(i+j))/((((i^2)+(j^2)))^s)),{i,j}]==-DirichletBeta[s]DirichletEta[s]

NDoubleSeries[((((-1))^(i+j))/((((i^2)+(j^2)))^2)),{i,j}]

-4DirichletBeta[2]DirichletEta[2]

-((Catalan (π^2))/3)//N

DoubleSeries[((((-1))^(i+j))/((((i^2)+(j^2)))^2)),{i,j}]

(∑_i^∞)(∑_j^∞)((((-1))^(i+j))/((((i^2)+(j^2)))^s))==η(2 s)-β(s) η(s)

Sum[((((-1))^(i+j))/((((i^2)+(j^2)))^s)),{i,∞},{j,∞}]

DirichletEta[2 s]-DirichletBeta[s] DirichletEta[s]//FullSimplify

Do[Print[{s,NSum[((((-1))^(i+j))/((((i^2)+(j^2)))^s)),{i,∞},{j,∞}]//Chop,DirichletEta[2s]-DirichletBeta[s]DirichletEta[s]//N}],{s,10}]

V6

Do[Print[{s,Sum[((((-1))^(i+j))/((((i^2)+(j^2)))^s)),{i,∞},{j,∞}]//Timing}],{s,3}]

V7

Do[Print[{s,Sum[((((-1))^(i+j))/((((i^2)+(j^2)))^s)),{i,∞},{j,∞}]//Timing}],{s,3}]

Double Series

(∑_i^∞)(∑_j^∞)(((i-1)! (j-1)!)/((i+j)!))==((π^2)/6)

Cc: "CLOITRE Benoit" <abcloitre@wanadoo.fr>
Date: Thu, 9 Dec 2004 14:18:21 +0100

V6

Sum[(((i-1)!(j-1)!)/((i+j)!)),{i,∞},{j,∞}]

Sum[(((i-1)!(j-1)!)/((i+j)!)),{j,∞},{i,∞}]

V7

Sum[(((i-1)!(j-1)!)/((i+j)!)),{i,∞},{j,∞}]//Timing

Sum[(((i-1)!(j-1)!)/((i+j)!)),{j,∞},{i,∞}]//Timing

(∑_(m=1)^∞)(∑_(n=1)^∞)(((m^2) n)/((3^m) (m (3^n)+(3^m) n)))==(9/32)

Borwein et al. 2004, p. 55

v5

NSum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,100},{n,100}]

NSum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,∞},{n,∞}]//Timing

Sum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,∞},{n,∞}]

v6

NSum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,100},{n,100}]

NSum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,∞},{n,∞}]//Timing

Sum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,∞},{n,∞}]

V7

Sum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{m,∞},{n,∞}]

Sum[(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),{n,∞},{m,∞}]

(1/2)Sum[((m n)/(3^(m+n))),{m,∞},{n,∞}]

Interchanging variables

{sum=(((m^2)n)/((3^m)(n (3^m)+m (3^n)))),sum/.{m->n,n->m}}

Simplify[Plus@@%]

(1/2)Sum[((m n)/(3^(m+n))),{m,∞},{n,∞}]

Sum[(m/(3^m)),{m,∞}]

(1/2)((Sum[(m/(3^m)),{m,∞}])^2)

(∑_m^∞)(∑_n^∞)(1/(2^(m n)))==1-((QPolyGamma3[0, 1, 1/2])/(log(2)))

Erdos-Borwein

V5

Sum[1/2^(m n),{m,∞},{n,∞}]

NSum[1/2^(m n),{m,∞},{n,∞}]

N[Plus@@Flatten[Table[1/2^(m n),{m,100},{n,100}]],20]//Timing

N[Plus@@Flatten[Table[1./2^(m n),{m,500},{n,500}]],20]//Timing

N[Plus@@Flatten[Table[1/2^(m n),{m,200},{n,200}]],20]//Timing

Sum[1/(2^n-1),{n,∞}]

NSum[1/(2^n-1),{n,∞},AccuracyGoal->∞,PrecisionGoal->200,WorkingPrecision->400]//Timing

N[Sum[1/(2^n-1),{n,800}],200]

V7

Sum[1/2^(m n),{m,∞},{n,∞}]//FullSimplify

Sum[1/2^(m n),{n,∞},{m,∞}]//FullSimplify

Sum[(1/((2^k)-1)),{k,∞}]//FullSimplify

(∑_m^∞)(∑_n^∞)(1/((m^2) n+m (n^2)+2 m n))==(7/4)

V6

Sum[(1/(((m^2)n+m (n^2)+2m n))),{m,∞},{n,∞}]

Sum[(1/(((m^2)n+m (n^2)+2m n))),{n,∞}]//FullSimplify

Sum[%,{m,∞}]

Sum[(1/(((m^2)n+m (n^2)+2m n))),{m,∞}]//FullSimplify

Sum[%,{n,∞}]

N[%]

V7

Sum[(1/(((m^2)n+m (n^2)+2m n))),{m,∞},{n,∞}]

Sum[(1/(((m^2)n+m (n^2)+2m n))),{n,∞},{m,∞}]

(∑_m^∞)(∑_n^∞)(1/((m^2) ((m^2)+m n+(n^2))))==X

V5

FullSimplify[Sum[(1/((m^2)(m^2+m n+n^2))),{m,∞}],n∈Integers&&n>0]

Sum[(1/((m^2)(m^2+m n+n^2))),{m,∞}]

N[Sum[(1/(6 (3 I+(Sqrt[3])) (n^3)))((3 I+(Sqrt[3])) n (π^2)-12 I HarmonicNumber[(1/2) (n-I (Sqrt[3]) n)]-6 (I+(Sqrt[3])) HarmonicNumber[(1/2) (n+I (Sqrt[3]) n)]),{n,1500}],20]

FullSimplify[Sum[(1/((m^2)(m^2+m n+n^2))),{n,∞}],m∈Integers&&m>0]

NSum[((I (HarmonicNumber[(1/2) (m-I (Sqrt[3]) m)]-HarmonicNumber[(1/2) (m+I (Sqrt[3]) m)]))/((Sqrt[3]) (m^3))),{m,∞},WorkingPrecision->20]

NSum[(1/((m^2)(m^2+m n+n^2))),{m,∞},{n,∞},WorkingPrecision->20]

RootReduce/@{(-1)^(1/3),-(-1)^(2/3)}

FullSimplify[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3)))]

Sum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,∞}]

Series[FunctionExpand[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3)))],{m,∞,20}]//Simplify

Normal[%23]

Plot[(3617/(8160 (m^19)))+(1/(12 (m^17)))-(1/(132 (m^13)))-(1/(240 (m^11)))+(1/(120 (m^7)))+(1/(12 (m^5)))-(1/(2 (m^4)))-(π/(3 (Sqrt[3]) (m^3)))+((I π Cot[(((-1))^(2/3)) m π])/((Sqrt[3]) (m^3))),{m,100,1000}]

Table[Cot[(((-1))^(2/3)) m π],{m,10}]//N

Normal[%]

(1/(12 (m^5)))-(1/(2 (m^4)))-(π/(3 (Sqrt[3]) (m^3)))+((I π Cot[(((-1))^(2/3)) m π])/((Sqrt[3]) (m^3)))/.m->2000.

((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3)))/.m->2000.

N[Sum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,2500}],110]//Timing

N[Sum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,3500}],110]//Timing

NSum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,∞},NSumTerms->1000,NSumExtraTerms->1000,WorkingPrecision->20]//Timing

NSum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,∞},NSumTerms->1000,NSumExtraTerms->1000,Method->Fit,WorkingPrecision->200]//Timing

NSum[((I (HarmonicNumber[-m(((-1))^(2/3))]-HarmonicNumber[m(((-1))^(1/3))]))/((Sqrt[3]) (m^3))),{m,∞},NSumTerms->1000,NSumExtraTerms->1000,WorkingPrecision->100]//Timing

NSum[(1/((m^2)(m^2+m n+n^2))),{m,∞},{n,∞},WorkingPrecision->20]

<<Graphics`Colors`

Plot[((I (HarmonicNumber[(1/2) (m-I (Sqrt[3]) m)]-HarmonicNumber[(1/2) (m+I (Sqrt[3]) m)]))/((Sqrt[3]) (m^3))),
{m,50,100},PlotStyle->Red];

terms=Table[(1/((Sqrt[3]) (m^3)))(I (HarmonicNumber[(1/2) (m-I (Sqrt[3]) m)]-HarmonicNumber[(1/2) (m+I (Sqrt[3]) m)])),{m,5000}];//Timing

N[Last[terms],20]

Plus@@N[terms,20]//Timing

Simplify[(1/2) (m-I (Sqrt[3]) m)-(1/2) (m+I (Sqrt[3]) m)]

N[FullSimplify[HarmonicNumber[2-I]-HarmonicNumber[2+I]],20]

NSum[(1/((Sqrt[3]) (m^3)))(I (HarmonicNumber[(1/2) (m-I (Sqrt[3]) m)]-HarmonicNumber[(1/2) (m+I (Sqrt[3]) m)])),{m,∞},WorkingPrecision->100,PrecisionGoal->20,AccuracyGoal->Infinity]

Factor[m/.Flatten[Solve[mm==m(1-I Sqrt[3])/2,m]]]

Factor[m/.Flatten[Solve[mm==m(1+I Sqrt[3])/2,m]]]

(1/(Sqrt[3]))I Pi^4/72((((((1-I Sqrt[3])/2))^3)/((((1+I Sqrt[3])/2))^3))-(((((1+I Sqrt[3])/2))^3)/((((1-I Sqrt[3])/2))^3)))

N[%]

Sum[HarmonicNumber[m]/(m^3),{m,(1-I Sqrt[3])/2,ComplexInfinity,(1-I Sqrt[3])/2}]

V7

TimeConstrained[Sum[(1/((m^2)(m^2+m n+n^2))),{m,∞},{n,∞}],300]//Timing

(∑_(n=1)^∞)((Subscript[H, n])/(n^3))==((π^4)/72)

V6

Sum[HarmonicNumber[n]/n^3,{n,∞}]

V6

Sum[HarmonicNumber[n]/n^3,{n,1,∞}]

(∑_j^∞)(∑_i^j)(i/j)==∞

Should give DirectedInfinity[1]

V6

Sum[i/j,{j,∞},{i,j}]

NSum[i/j,{j,1.,∞},{i,1.,j}]

V7

Sum[i/j,{j,∞},{i,j}]//Timing

(∑_i^∞)(∑_j^∞)(1/((i^2)+(j^2)))

Numeric

NSum[1/(i^2+j^2),{i,∞},{j,∞}]

V6

After a lot of messages this gives the right result..

Sum[1/(i^2+j^2),{i,∞},{j,∞}]

NSum[1/(i^2+j^2),{i,∞},{j,∞}]

V7

TimeConstrained[Sum[1/(i^2+j^2),{i,∞},{j,∞}],300]//Timing

(∑_i^∞)(∑_j^∞)(1/(((i+j))^2))==∞

V6

Long Output. The result is (a very big) Real, should be DirectedInfinity[1] because (1/(((i+j))^2))==(1/(((i^2)+2 i j+(j^2))))(consider fixed i)

Sum[(1/(((i+j))^2)),{i,∞},{j,∞}]

NSum[(1/(((i+j))^2)),{i,∞},{j,∞}]

V7

TimeConstrained[Sum[(1/(((i+j))^2)),{i,∞},{j,∞}],300]//Timing

(∑_i^∞)(∑_j^∞)(1/((2^i)+(2^j)))

Numeric

NSum[(1/((2^i)+(2^j))),{i,∞},{j,∞}]

V6

Sum[(1/((2^i)+(2^j))),{i,∞},{j,∞}]

NSum[1/(2^i+2^j),{i,∞},{j,∞}]

Sum[(1/((2^i)*(2^j))),{i,∞},{j,∞}]

After some messages this gives the right result

NSum[1/(2^i*2^j),{i,∞},{j,∞}]

Sum[1/i^2,{i,∞}]

Sum[1/(i*j)^2,{i,∞},{j,∞}]

V7

TimeConstrained[Sum[(1/((2^i)+(2^j))),{i,∞},{j,∞}],300]//Timing

(∑_i^∞)(∑_j^∞)(1/(((i j))^2))==((π^4)/36)

V6

After a lot of messages this evaluatse to the right result

NSum[1/(i*j)^2,{i,∞},{j,∞}]

2.705808083987627`

Sum Check

Sum[1/(i*j)^2,{i,∞},{j,∞}]
%//N//FullForm

V7

Sum[1/(i*j)^2,{i,∞},{j,∞}]

(∑_i^∞)(∑_j^∞)(1/((i^2)+(2^(i+j))))

Numeric

NSum[1/(i^2+2^(i+j)),{i,∞},{j,∞}]

V6

Sum[1/(i^2+2^(i+j)),{i,∞},{j,∞}]

After some output, and a lot of time this gives the right result

NSum[1/(i^2+2^(i+j)),{i,∞},{j,∞}]//FullForm

0.8157831160216186`

Finite check

sum=0;
Do[sum+=1./(i^2+2^(i+j)),{i,1,10},{j,1,10}]
sum//FullForm

sum=0;
Do[sum+=1./(i^2+2^(i+j)),{i,1,100},{j,1,100}]
sum//FullForm

sum=0;
Do[sum+=1./(i^2+2^(i+j)),{i,1,500},{j,1,500}]
sum//FullForm

sum=0;
Do[sum+=1./(i^2+2^(i+j)),{i,1,1000},{j,1,1000}]
sum//FullForm

V7

TimeConstrained[Sum[1/(i^2+2^(i+j)),{i,∞},{j,∞}],300]//Timing

TimeConstrained[Sum[1/(i^2+2^(i+j)),{j,∞},{i,∞}],300]//Timing

(∑_i^∞)(∑_j^∞)((i+i j)/(i+(j^2)))

V6

Sum[((i+j I)/(i+(j^2))),{i,∞},{j,∞}]

NSum[(i+j*I)/(i+j^2),{i,∞},{j,∞}]

V7

TimeConstrained[Sum[((i+j I)/(i+(j^2))),{i,∞},{j,∞}],300]//Timing