2
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It looks like this.

Limit[Sum[1/((2 n + 1) Pi I - x), {n, -m, m}], m -> ∞]
= -(1/2) Tanh[x/2]

Sum[1/((2 n + 1) Pi I - x), {n, -∞, ∞}]
Sum::div: Sum does not converge.

Turning on Regularization options does not help to converge. What is the problem here?

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  • $\begingroup$ Try: Sum[Re[1/(I (1 + 2 n) π - x)] // ComplexExpand, {n, -∞, ∞}] $\endgroup$ – Mariusz Iwaniuk Nov 20 '17 at 12:56
  • 1
    $\begingroup$ Another workaround: (1/((2 n + 1) Pi I - x) /. n -> 0) + Sum[1/((2 n + 1) Pi I - x) + (1/((2 n + 1) Pi I - x) /. n -> -n), {n, 1, Infinity}] // Simplify $\endgroup$ – Bob Hanlon Nov 20 '17 at 13:31
  • $\begingroup$ I was going to refer to mathematica.stackexchange.com/a/133024, but SumConvergence only seems to check convergence for "one-sided" sums, which are divergent in this case. (One can still try. For Direction->1 and Direction->-1 the function SumConvergence applied to 1/((2 n + 1) Pi I - x) returns unevaluated, except for Method -> "DivergenceTest", where it gives True. Also, SumConvergence gives the same results for the combination1/((2 n + 1) Pi I - x) + (1/((2 n + 1) Pi I - x) /. n -> -n) from @BobHanlon's comment, whose infinite sum Sum does compute. Strange.) $\endgroup$ – Jules Lamers Nov 21 '17 at 18:22

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