Solve for $C_k$
$$\sum _{k=-\infty }^{\infty } \frac{i (-1)^k e^{i k t}}{k}=4 \sum _{k=-\infty }^{\infty } k^2 C_k \left(-e^{i k t}\right)+4 \sum _{k=-\infty }^{\infty } i k C_k e^{i k t}+17 \sum _{k=-\infty }^{\infty } C_k e^{i k t}$$
The first thing I noticed was that both LHS and RHS has the factor $e^{ikt}$ attached in their respective summation terms. Therefore we can safely remove this from our equation:
$$\sum _{k=-\infty }^{\infty } \frac{i (-1)^k}{k}=-4 \sum _{k=-\infty }^{\infty } k^2 C_k+4 i \sum _{k=-\infty }^{\infty } k C_k+17 \sum _{k=-\infty }^{\infty } C_k$$
I'm stuck here, any idea what I can do next to solve for $C_k$. I'm not too familiar with summation marks, and what rules there are to them, but I'd like to cancel them out if possible. Notice they're all in the same range, $[-\infty,\infty]$.
You can't just cancel a term like that! Thats because the $\Sigma$ is more like a function than a multiplier, and just because say $\sin(\cos(x) x) = \sin(\cos (x) 2x)$ for some $x$, doesn't mean you can conclude that $\sin(x) = \sin(2x)$ for this same $x$. For further confirmation, try it for a small sum. $\sum_{k=1}^2 e^{ikt} = \sum_{k=1}^2 k C_k e^{ikt}$
What you can do, is compare coefficients of each $e^{ikt}$. This is allowed because they form a basis.
So we have
$$\frac{i (-1)^k e^{i k t}}{k}=4 k^2 C_k \left(-e^{i k t}\right)+4 i k C_k e^{i k t}+17 C_k e^{i k t}$$
Now you can cancel the $e^{ikt}$s, and then rearrange for $C_k$.
(edit- Ron Gordon's comment raises an important point, but the method I have lined out above is still correct)