I calculated a complex Fourier series and got this result: $$ \sum_{k=-n}^{n}\frac{-(-1)^k -1}{(k^2-1)}e^{ikt}. $$
However, since this is the Fourier series of an absolute value function, I was told I needed to define the series both for $+(k^2-1)$ and $-(k^2-1)$. Moreover, I'm confused about what to do about the case when $k = +1$ or $-1$, since that makes the denominator $0$.
I am extremely new to this difficult topic so any help is appreciated, and sorry if I was unclear.
The formula used for the coefficients: $c_k = \frac{1}{\pi} \int_0^\pi f(t)e^{-ikt}\,dt$.
The general formula you have for the coefficients is correct (i.e. $c_k = -\frac{(-1)^k+1}{k^2-1}$). As for the $k=\pm 1$ case, since there is (a priori) a problem with division by zero, the solution is to just solve the relevant integrals by hand, using the fact that $|\sin(t)|$ is nonnegative on $[0,\pi]$: $$ c_{\pm1} = \frac{1}{\pi}\int_0^\pi |\sin(t)|e^{\mp it} dt = \frac1\pi \int_0^\pi\sin(t)e^{\mp it}dt = \frac1\pi\left.\left(-\frac{e^{\mp2 it}}4 \mp \frac{it}{2}\right)\right|_{t=0}^\pi = \mp\frac i2. $$ When you evaluate the sum for different values of $n$, you'll see that the $c_{\pm1}$ contributions cancel. But this corresponds to the odd terms vanishing in the general formula, so it may be helpful to reindex the general series so that you're only summing over the even terms.