NSolve[(-(2/Cos[Pi*x]))*Sum[(-1)^i*(2*Pi*x)^(2*i)*Sum[(-1)^j/((2*Pi)^(2*j)*Zeta[2*j]*(2*i+1-2*j)!),{j,1,i}],{i,1,Infinity}]0,x](*Linecopiedforyourpost*)
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NSolve-2Sec[πx]0,x
∞
∑
i=1
i
(-1)
2i
(2π)
2i
x
i
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j=1
j
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-2j
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(1+2i-2j)!Zeta[2j]
SumConvergence,j(*SumConvergencetestdoesnotworkforthelastsum.Mycomputermaynothaveenoughresourcestoevaluatethetestortheexpressionmayhaveerrors*)SumConvergence[,i](*Ihadtotestthefirstpartwithoutthenestedsumsincethenestedsumdidnotpasstheconvergencetest.*)
j
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(1+2i-2j)!Zeta[2j]
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SumConvergence,j
j
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(1+2i-2j)!Zeta[2j]
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Abs[x]<||2πx1||1+2πx0
1
2π
(*SumConvergenceofthefirstsumshowsthatitonlyconvergeswhen-<x<*)
1
2π
1
2π
f[x_,n_]=(-(2/Cos[Pi*x]))*Sum[(-1)^i*(2*Pi*x)^(2*i)*Sum[(-1)^j/((2*Pi)^(2*j)*Zeta[2*j]*(2*i+1-2*j)!),{j,1,i}],{i,1,n}](*Mycomputerwillnotevaluatetheinfinitesum.Therefore,Ireplace"Infinity"withn.Then,Iplottedthefunctiontoshowwhathappensasngetsverylarge.*)Plot3Df[x,n],x,,,{n,0,30},AxesLabel{"x","n","f(x)"}
-1
2π
1
2π
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-2Sec[πx]
n
∑
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(1+2i-2j)!Zeta[2j]
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(*Thefunctionseemstoconvergefastasnbecomeslarger.Thisgraphisstillonlyanapproximation,andwithitImakethebestassumptionthattheonlytimethefullequationholdstrueiswhenx=0.*)