 Solution

The literal expression in the original PDF is undefined. Indeed, hyperbolic tangent satisfies $\tanh w=\sinh w/\cosh w$, andAlready the $n=2$ summand contains $\tanh (i\pi /2)$, a pole. The shifted term has the same issue when $n\equiv 3(\mathrm{mod}4)$. Thus this is not a finite sum of defined complex numbers, and it has no literal real or imaginary part. Cancelling two undefined terms is not ordinary arithmetic.

There is nevertheless a natural interpretation by meromorphic telescoping cancellation. Insert the same small positive real shift $\epsilon$ into every hyperbolic tangent argument. Each term is then finite, and the telescoping sum isSince hyperbolic tangent has period $i\pi$, its first argument can be reduced to $\epsilon +i\pi /4$. Taking the limit givesThis is a common-shift continuation of the sum, plausibly the intended telescoping answer. It requires that additional convention; it is not the value of the expression as printed.