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After some long derivations with a certain 1D Hamiltonian, I end up with the following sum $$\sum_{n=2}^\infty \frac{1}{e^{an^2(1+\frac{b}{4n^2-1})}-1}$$ But I am completely stumped by the $(1+\frac{b}{4n^2-1})$ term. Is there any trick I can use to compute this sum?

I am interested in the behavior of the sum with respect to $a$

$a$ and $b$ are positive constants.

DarkBulle
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    Similar questions were asked and answered a lot. Try numerically: f[a_?NumericQ, b_?NumericQ] := NSum[1/(Exp[a*n^2*(1 + b/(4 n^2 - 1))] - 1), {n, 2, Infinity}], then e.g. f[0.4, 0.5] results in 0.264993. – user64494 Mar 16 '22 at 11:59
  • thank you for your comment @user64494 , I was in fact wondering If I could get an analytical result, to say physical things about my problem, I am interested in the behavior of the sum with respect to the $a$ parameter – DarkBulle Mar 16 '22 at 12:01
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    Do you have reason to believe that an analytical result is possible? – bbgodfrey Mar 16 '22 at 13:41
  • @bbgodfrey, absolutely not – DarkBulle Mar 16 '22 at 13:42
  • Given that even the much more simple $$\sum_{n=1}^\infty \frac{1}{e^{n^2}}$$ does not have an analytical result, I don't see much hope in the posted sum having any analytical result. – 5xum Mar 17 '22 at 07:20

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