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What is the correct and conventional way to express the Dirichlet eta function as Dirichlet characters in Mathematica?

Table[(2*DirichletCharacter[2, 1, n] - 1)/n^s, {n, 1, Infinity}]

or only the numerators as a finite sequence:

Table[(2*DirichletCharacter[2, 1, n] - 1), {n, 1, 12}]

{1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1}

https://en.wikipedia.org/wiki/Dirichlet_character#Modulus_2

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I'm not sure if this is what you're looking for but it is easy to check in Mathematica that the infinite sum using DirichletCharacter[]

Sum[(2 DirichletCharacter[2, 1, n] - 1)/n^s, {n, 1, \[Infinity]}]

(* Out[38]= 2^-s (-2 + 2^s) Zeta[s] *)

is indeed DirichletEta[s]

% == DirichletEta[s] // FullSimplify

(* Out[37]= True *)
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  • $\begingroup$ Yes I see.I have another question about this though: math.stackexchange.com/questions/1895070/… $\endgroup$ – Mats Granvik Aug 17 '16 at 14:29
  • $\begingroup$ But thanks for the effort. Not directly related either but I am seeking to apply the symmetric functional equation here: lmfdb.org/L/degree1 $\endgroup$ – Mats Granvik Aug 17 '16 at 14:31
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    $\begingroup$ Another way to express the first relation: 2 DirichletL[2, 1, s] - DirichletL[1, 1, s] == DirichletEta[s] // FullSimplify. $\endgroup$ – J. M. is away Aug 17 '16 at 17:08
  • $\begingroup$ As to your first comment: why not check the validity of the two formulas in Mathematica, at least numerically? $\endgroup$ – Dr. Wolfgang Hintze Aug 18 '16 at 6:37

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