= Solution
The literal expression in the original PDF is undefined. Indeed, <hyperbolic tangent> satisfies $\tanh w=\sinh w/\cosh w$, and
$$
\cosh(in\pi/4)=\cos(n\pi/4)=0\quad\text{when }n\equiv2\pmod4.
$$
Already the $n=2$ summand contains $\tanh(i\pi/2)$, a <pole>. The shifted term has the same issue when $n\equiv3\pmod4$. Thus this is not a finite sum of defined <complex numbers>, and it has \b[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 $\varepsilon$ into every <hyperbolic tangent> argument. Each term is then finite, and the <telescoping sum> is
$$
S(\varepsilon)=\tanh(\varepsilon+121i\pi/4)-\tanh\varepsilon.
$$
Since <hyperbolic tangent> has period $i\pi$, its first argument can be reduced to $\varepsilon+i\pi/4$. Taking the <limit> gives
$$
\boxed{\lim_{\varepsilon\to0^+}S(\varepsilon)=i,\qquad\Re=0,\quad\Im=1.}
$$
This 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.
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