I'm actually dealing with the fractional part function,
$$\{x\} = x - \lfloor x\rfloor$$
On Wikipedia, a Fourier series expansion is given as,
$$\{x\} = \frac{1}{2} - \frac{1}{\pi}\sum_{k=1}^\infty \frac{\sin(2\pi k x)}{k}$$
However, since $\sin(2\pi n)$ is $0$, we can actually just substitute the fractional part of x on the inside as well.
$$\{x\} = \frac{1}{2} - \frac{1}{\pi}\sum_{k=1}^\infty \frac{\sin(2\pi k \{x\})}{k}$$
This gives us a recursive definition for the fractional part, as we could just substitute the whole expression inside again,
$$\{x\} = \frac{1}{2} - \frac{1}{\pi}\sum_{k=1}^\infty \frac{\sin\left(2\pi k \left(\frac{1}{2} - \frac{1}{\pi}\sum_{m=1}^\infty\frac{\sin(2\pi m \ ...)}{m}\right)\right)}{k}$$
Is there not some kind of trick to use this fact to get a closed-form expression of $\{x\}$ without summing the original infinite series?