Series

Author

Eric W. Weisstein
December 26, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Calculus/Series.nb.

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

©2008 Wolfram Research, Inc. except for portions noted otherwise

Borwein et al. Experimental Math in Action, Ch. 3 #4

<<Utilities`Simplify`

(∑_(n=0)^∞)((50 n-6)/((2^n) (({{3 n}, {n}}))))==π

V6.0

Sum[((50n-6)/((2^n)Binomial[3n,n])),{n,0,∞}]

TranscendentalRecognize[N[%,1000],{π}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

$Version

Sum[((50n-6)/((2^n)Binomial[3n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{π}]

(∑_(n=0)^∞)((2^(n+1))/((({{2 n}, {n}}))))==4+π

V6

Sum[((2^(n+1))/(Binomial[2n,n])),{n,0,∞}]

V7

Sum[((2^(n+1))/(Binomial[2n,n])),{n,0,∞}]

(∑_(n=0)^∞)((12 n (2^(2 n)))/((({{4 n}, {2 n}}))))==3 π+8

Sum[((12n (2^(2n)))/(Binomial[4n,2n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{π}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

Sum[((12n (2^(2n)))/(Binomial[4n,2n])),{n,0,∞}]

TranscendentalRecognize[N[%,1000],{π}]

(∑_(n=0)^∞)(((4 n)! (8 n+1))/((4^(4 n)) (((n!))^4)))==(2/(π (Sqrt[3])))?

V6

Sum[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,∞}]

(2/(π(Sqrt[3])))//N

Table[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,20}]//N

SequenceLimit[FoldList[Plus,0,N[Table[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,20}],1000]]]

N[Sum[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,100}],20]

NSum[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,∞}]

V7

Sum[(((4n)!(8n+1))/((4^(4n))((n!)^4))),{n,0,∞}]

SumConvergence[(((4n)!(8n+1))/((4^(4n))((n!)^4))),n]

(∑_(n=0)^∞)(((4 n)! (280 n+19))/((4^(4 n)) (((n!))^4) (99^(2 n+1))))==(2/(π (Sqrt[11])))

V6

Sum[(((4n)!(19+280n))/((4^(4n))((n!)^4)(99^(2n+1)))),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[(1/%)^2,1000],{π,π^2}]

Developer`ZeroQ[Subtract@@{(2/(π(Sqrt[11]))),%%}]

V7

Sum[(((4n)!(19+280n))/((4^(4n))((n!)^4)(99^(2n+1)))),{n,0,∞}]

(∑_(n=0)^∞)(((2 n)! (3 n)! (4^n) (33 n+4))/((((n!))^5) (108^n) (125^n)))==((15 (Sqrt[3]))/(2 π))

V6

Sum[(((2n)!(3n)!(4^n)(4+33n))/(((n!)^5)(108^n)(125^n))),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[(1/%)^2,1000],{π,π^2}]

Developer`ZeroQ[Subtract@@{%%,((15 (Sqrt[3]))/(2 π))}]

V7

Sum[(((2n)!(3n)!(4^n)(4+33n))/(((n!)^5)(108^n)(125^n))),{n,0,∞}]

TranscendentalRecognize[N[(1/%)^2,1000],{π,π^2}]

(∑_(n=0)^∞)(((((-27))^n) (90 n+177))/((16^n) (({{3 n}, {n}}))))==120-64 log(2)

Sum[(((((-27))^n)(90n+177))/((16^n)Binomial[3n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{Log[2]}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

Sum[(((((-27))^n)(90n+177))/((16^n)Binomial[3n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{Log[2]}]

(∑_(n=0)^∞)((275 n-158)/((2^n) (({{3 n}, {n}}))))==6 log(2)-135

V6

Sum[((275n-158)/((2^n)Binomial[3n,n])),{n,0,∞}]

TranscendentalRecognize[N[%,1000],{Log[2]}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

Sum[((275n-158)/((2^n)Binomial[3n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{Log[2]}]

(∑_(n=0)^∞)(((8^n) (-430 (n^2)+6240 n+520))/((({{4 n}, {n}}))))==-45 π-644

V6

$Version

Sum[(((8^n)(520+6240n-430(n^2)))/(Binomial[4n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{π}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

$Version

Sum[(((8^n)(520+6240n-430(n^2)))/(Binomial[4n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{π}]

(∑_(n=1)^∞)(((({{2 n}, {n}})))/((n^2) (4^n)))==(1/6) ((π^2)-12 (log^2)(2))

Sum[((Binomial[2n,n])/((n^2)(4^n))),{n,∞}]//LogExpand

V7

Sum[((Binomial[2n,n])/((n^2)(4^n))),{n,∞}]//LogExpand

(∑_(n=1)^∞)((((-1))^n)/((n^3) (2^n) (({{2 n}, {n}}))))==(((log^3)(2))/6)-((ζ(3))/4)

Sum[((((-1))^n)/((n^3)(2^n)Binomial[2n,n])),{n,1,∞}]//FullSimplify

V7

Sum[((((-1))^n)/((n^3)(2^n)Binomial[2n,n])),{n,∞}]//FullSimplify

(∑_(n=0)^∞)(((8^n) (338-245 n))/((3^n) (({{3 n}, {n}}))))==-18 log(3)-6 (Sqrt[3]) π+162

Sum[(((8^n)(338-245n))/((3^n)Binomial[3n,n])),{n,0,∞}]//FullSimplify

TranscendentalRecognize[N[%,1000],{π (Sqrt[3]),Log[3]}]

Developer`ZeroQ[Subtract@@{%,%%}]

V7

Sum[(((8^n)(338-245n))/((3^n)Binomial[3n,n])),{n,0,∞}]

TranscendentalRecognize[N[%,1000],{π (Sqrt[3]),Log[3]}]

(∑_(n=1)^∞)(((((-9))^n) (Binomial[2 n, n]))/(6 (n^2) (64^n)))-(∑_(n=1)^∞)(((3^n) (Binomial[2 n, n]))/((n^2) (16^n)))==-2 (PolyLog[2, (1/4)])+((PolyLog[2, -(1/8)])/3)+((5 (log^2)(2))/2)+(((log^2)(3))/3)-2 log(3) log(2)

<<Utilities`Simplify`

Sum[(((((-9))^n)Binomial[2n,n])/(6(n^2)(64^n))),{n,∞}]-Sum[(((3^n)Binomial[2n,n])/((n^2)(16^n))),{n,∞}]//FullSimplify//LogExpand//LogExpand

N[%2,20]

NSum[(((((-9))^n)Binomial[2n,n])/(6(n^2)(64^n))),{n,∞}]-NSum[(((3^n)Binomial[2n,n])/((n^2)(16^n))),{n,∞}]

V7

Sum[(((((-9))^n)Binomial[2n,n])/(6(n^2)(64^n))),{n,∞}]-Sum[(((3^n)Binomial[2n,n])/((n^2)(16^n))),{n,∞}]

FullSimplify[%]//LogExpand//Expand

FullSimplify[%]

Borwein et al. Experimental Math in Action, Ch. 9 Problems

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

Factor//@FullSimplify[Sum[(1/(2j-1))(x^(2k-1)),{k,∞},{j,k}],0<x<1]

Factor//@FullSimplify[Sum[(1/(2j-1))((x^(2k))/k),{k,∞},{j,k}],0<x<1]

V6

Factor//@FullSimplify[Sum[(1/(2j-1))((x^(2k))/(k^2)),{k,∞},{j,k}],0<x<1]

V7

FullSimplify[Sum[(1/(2j-1))((x^(2k))/(k^2)),{k,∞},{j,k}],0<x<1]//Timing

Misc

(∑_(n=0)^∞)((((1/2)))^n)==2

Sum[((((1/2)))^n),{n,0,∞}]

(∑_(k=1, 3, …)^∞)(((e^(-k x)) sin(k y))/k)== (1/2)(tan^(-1))(((sin(y))/(sinh(x))))

V5

Timing[s=Sum[Exp[-k x]Sin[k y]/k,{k,1,Infinity,2}]]

FullSimplify[s]

Show[GraphicsArray[{Block[{$DisplayFunction=Identity},
Plot3D[Evaluate[#[s-1/2ArcTan[Sin[y]/Sinh[x]]]],{x,-3,3},{y,-3,3},AxesLabel->TraditionalForm/@{"x","y","z"}]&/@{Re,Im}
]}]]

Show[GraphicsArray[{Block[{$DisplayFunction=Identity},
Plot3D[Evaluate[#[s-1/2ArcTan[Sinh[x],Sin[y]]]],{x,-3,3},{y,-3,3},AxesLabel->TraditionalForm/@{"x","y","z"}]&/@{Re,Im}
]}]]

Expand[TrigToExp[Subtract@@{(1/2) I (-ArcCoth[(E^(x-I y))]+ArcCoth[(E^(x+I y))]),1/2ArcTan[Sin[y]/Sinh[x]]}]4/I]

FullSimplify[(1/4)I Log[ExpandAll[(1/(1-(E^(-x-I y))))(1+(E^(-x-I y)))(1-(E^(-x+I y)))(1/(1+(E^(-x+I y))))(1-(((E^(-I y))-(E^(I y)))/(-(E^(-x))+(E^x))))(1/(1+(((E^(-I y))-(E^(I y)))/(-(E^(-x))+(E^x)))))]],(x|y)∈Reals]

V7

Sum[Exp[-k x]Sin[k y]/k,{k,1,Infinity,2}]

(∑_n^∞)((csch(n x) cosh(n y))/n)==ln((((Subscript[ϑ, 1])((π/3),exp(-(x/3))))/((Subscript[ϑ, 4])(((i y)/2),exp(-x)))))-ln((Sqrt[3]))+(x/12)

math-fun
17 Jul 2008

V7

Sum[((Cosh[n y])/(n Sinh[n x])),{n,∞}]//Timing

Subtract@@@DeleteCases[Table[
Block[{x=RandomReal[{0,1},WorkingPrecision->20],y=RandomReal[{0,1},WorkingPrecision->20]},
{NSum[((Cosh[n y])/(n Sinh[n x])),{n,∞}],Log[((EllipticTheta[1,π/3,Exp[-x/3]])/(EllipticTheta[4,I y/2,Exp[-x]]))]+x/12-Log[Sqrt[3]]}
],{10}],{ComplexInfinity,_}]//Chop[#,10^-5]&//Union//Quiet

Sum[((Cosh[n y])/(n Sinh[n x])),{n,∞}]/.{x->1/3,y->1/6}

With[{x=1/3,y=1/6},Sum[((Cosh[n y])/(n Sinh[n x])),{n,∞}]]

With[{x=1/3,y=1/6},NSum[((Cosh[n y])/(n Sinh[n x])),{n,∞}]]

With[{x=1/3,y=1/6},Log[((EllipticTheta[1,π/3,Exp[-x/3]])/(EllipticTheta[4,I y/2,Exp[-x]]))]+x/12-Log[Sqrt[3]]]

(1/36)-((Log[3])/2)+Log[((EllipticTheta[1,(π/3),(1/(E^(1/9)))])/(EllipticTheta[4,(I/12),(1/(E^(1/3)))]))]//N

(∑_(n=0)^∞)(((((-1))^n) (3 n)! (x^(2 n)))/(((n! (2 n)!))^3))==((∑_(n=0)^∞)((x^n)/(((n!))^3)))( (∑_(n=0)^∞)(((((-1))^n) (x^n))/(((n!))^3)))

Ramanujan's Identity

V5

TraditionalForm[s1=Sum[(((-1))^n)(((3n)!)/(((n!(2n)!))^3))(x^(2n)),{n,0,∞}]]

TraditionalForm[s2=Sum[((x^n)/(((n!))^3)),{n,0,∞}]Sum[(((-1))^n)((x^n)/(((n!))^3)),{n,0,∞}]]

FullSimplify[FunctionExpand[s1-s2]]

V7

TraditionalForm[s1=Sum[(((-1))^n)(((3n)!)/(((n!(2n)!))^3))(x^(2n)),{n,0,∞}]]

TraditionalForm[s2=Sum[((x^n)/(((n!))^3)),{n,0,∞}]Sum[(((-1))^n)((x^n)/(((n!))^3)),{n,0,∞}]]

Table[N[s1-s2/.x->RandomComplex[5(I+1){-1,1},WorkingPrecision->20]],{100}]//Chop//Union

PossibleZeroQ[-HypergeometricPFQ[{},{1,1},-x] HypergeometricPFQ[{},{1,1},x]+HypergeometricPFQ[{(1/3),(2/3)},{(1/2),(1/2),(1/2),1,1,1,1},-((27 (x^2))/64)]]

((((∑_(k=0)^∞)(x^k)))^p)==((((1/(1-x))))^p)

((Sum[(x^k),{k,0,∞}])^p)

(1/((p-1)!))(∑_(n=0)^∞)(((x^n) (n+p-1)!)/(n!))==(((1-x))^(-p))

(1/((p-1)!))Sum[(((n+p-1)!)/(n!))(x^n),{n,0,∞}]

(1/((p-1)!))(∑_(n=0)^∞)(x^n) (Pochhammer[n+1, p-1])==(((1-x))^(-p))

(1/((p-1)!))Sum[Pochhammer[n+1,p-1](x^n),{n,0,∞}]

(∑_(n=1)^∞)((cos((9/((Sqrt[(π^2) (n^2)-9])+π n))))/(n^2))==-((π^2)/(12 (e^3)))

FullSimplify[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],n∈Integers&&n>0]

Subtract@@@Table[{(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))]},{n,10}]//N//Chop//Union

v5.0

NSum[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],{n,∞},WorkingPrecision->20]//Timing

N[-((π^2)/(12(E^3))),20]

NSum[(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),{n,∞}]//Timing

Sum[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],{n,∞}]

Sum[(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),{n,∞}]

v5.1

NSum[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],{n,∞}]//Timing

NSum[(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),{n,∞}]//Timing

Sum[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],{n,∞}]

Sum[(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),{n,∞}]

V7

Sum[(1/(n^2))Cos[(9/(n π+(Sqrt[(n^2)(π^2)-9])))],{n,∞}]

N[%]

(∑_n^∞)(((((-1))^n) cos((Sqrt[(π^2) (n^2)-9])))/(n^2))==-((π^2)/(12 (e^3)))

Sum[(((((-1))^n) Cos[(Sqrt[-9+(n^2) (π^2)])])/(n^2)),{n,∞}]

N[%]

(∑_k^∞)(1/((((π^2) (k^2)+1))^n))

Sum[(1/((1+k^2Pi^2)^n)),{k,∞}]

p=Together/@TrigToExp[Table[(n-1)!2^n(E^2-1)^n Sum[(1/((1+k^2Pi^2)^n)),{k,∞}],{n,8}]]

Reverse/@(CoefficientList[#,E^2]&/@p)

(∑_k^∞)(1/((((π^2) (((k+(1/2)))^2)+1))^n))

Sum[(1/((1+(k+1/2)^2Pi^2)^n)),{k,∞}]

q=FullSimplify[TrigToExp[Table[(Sum[(1/((1+(k+1/2)^2Pi^2)^n)),{k,∞}]+((4^n)/((4+Pi^2)^n)))(n-1)!2^n(E^2+1)^n,{n,8}]]
]

Reverse/@(CoefficientList[#,E^2]&/@q)