# Summation returns unevaluated

Sum[E^(-I a n)/(1 + E^(-b n)), {n, -\[Infinity], \[Infinity]},
Assumptions -> {Im[a] < 0, b > 0}]


Mathematica returns the same expression,because dosen't know the answer.

This is only analytics approximations:

HoldForm[Sum[
E^(-I a n)/(1 + E^(-b n)), {n, -\[Infinity], \[Infinity]}] ==
Sum[E^(-a n)/(1 + E^(-b n)), {n, -Infinity, Infinity}] ==
Limit[E^(-a n)/(1 + E^(-b n)), n -> 0] +
Sum[E^(-a n)/(1 + E^(-b n)) + (E^(-a n)/(1 + E^(-b n)) /.
n -> -n), {n, 1, Infinity}] ==
1/2 + Sum[
E^(-a n)/(1 + E^(-b n)) + E^(a n)/(1 + E^(b n)), {n, 1,
Infinity}] ==
1/2 + Sum[
Cosh[n (a - b/2)] Sech[(b n)/2], {n, 1, Infinity}]] // TeXForm


$$\sum _{n=-\infty }^{\infty } \frac{e^{-i a n}}{1+e^{-b n}}=\\\sum _{n=-\infty }^{\infty } \frac{e^{-a n}}{1+e^{-b n}}=\\\underset{n\to 0}{\text{lim}}\frac{e^{-a n}}{1+e^{-b n}}+\sum _{n=1}^{\infty } \left(\frac{e^{-a n}}{1+e^{-b n}}+\left(\frac{e^{-a n}}{1+e^{-b n}}\text{/.}\, n\to -n\right)\right)=\\\frac{1}{2}+\sum _{n=1}^{\infty } \left(\frac{e^{-a n}}{1+e^{-b n}}+\frac{e^{a n}}{1+e^{b n}}\right)=\\\frac{1}{2}+\sum _{n=1}^{\infty } \cosh \left(n \left(a-\frac{b}{2}\right)\right) \text{sech}\left(\frac{b n}{2}\right)$$

Using Abel-Plana formula:

We can use formula if limit is equal to zero:

u[n_] := Cosh[n*(a - b/2)] Sech[(b n)/2];

Limit[Exp[-2 Pi Abs[y]]*Abs[u[x + I y]], y -> Infinity,
Assumptions -> {b > 0, 0 < a < 1, 0 < a < b, x \[Element] Reals}]
(* 0 *)(* OK.On page: 9->(2.17) *)

1/2 - Limit[u[n], n -> 0] + Integrate[u[x], {x, 0, Infinity}, Assumptions -> {b > 0, a < 1, 0 < a < b}]

(* (π Csc[(a π)/b])/b *)

I*(u[I t] - u[-I t])/(Exp[2 Pi*t] - 1) // Simplify
(*The third part of the Abel-Plana formula is zero !!! *)
(* 0 *)


Then we have:

$$\color{red}{\sum _{n=-\infty }^{\infty } \frac{e^{-i a n}}{1+e^{-b n}}}\approx \frac{\pi \csc \left(\frac{a \pi }{b}\right)}{b}\approx \color{red}{\frac{\pi \csc \left(\frac{i a \pi }{b}\right)}{b}}$$

for:$-1<\Im(a)<0$ and $b>0$.

Check:

  a = -1/10 I;
b = 1;

s1=N[(π Csc[(I a π)/b])/b , 20]
(* 10.166407384630519632 *)
s2=NSum[E^(-I a n)/(1 + E^(-b n)) , {n, -Infinity, Infinity}, WorkingPrecision -> 20,
NSumTerms -> 10000, Method -> "AlternatingSigns"]
(* 10.166407395019257050 *)

Abs[(s2 - s1)/s2](*Absolute error *)

(* 1.0218690846*10^-9 *)

• Wow! I am impressed.. Commented Jun 16, 2018 at 9:14
• But replacing 'a' with 'a + 2 pi' should give the same answer. Commented Jun 16, 2018 at 10:08
• @QuasarSupernova. It was also possible to use: Integrate[E^(-I*a n)/(1 + E^(-b n)), {n, -Infinity, Infinity}, Assumptions -> {b > 0, Im[a] < 0}] and we have:-((I \[Pi] Csch[(a \[Pi])/b])/b)= (\[Pi] Csc[(I a \[Pi])/b]) well, its not a exact answer,only approximation.If b>>1 then absolute error increases. Commented Jun 16, 2018 at 10:21
• Sorry, I want an exact answer. I find it hard to believe that it cannot be expressed in terms of functions known to Mathematica like MeijerG function. Commented Jun 17, 2018 at 10:46
• I also want many things?MMA know about MeijerG reference.wolfram.com/language/ref/MeijerG.html Commented Jun 17, 2018 at 11:01