= Solution
Over one period, the graph rises linearly from the limiting point $(-\pi,0)$ to $(0,\pi)$, then drops to zero and stays there up to $\pi$. Repeat this pattern every $2\pi$. The given open endpoints do not fix values at the jumps, but changing individual point values does not change any <Fourier coefficient>.
Only the negative half-interval contributes. The constant coefficient is
$$
a_0=\frac1\pi\int_{-\pi}^0(x+\pi)dx=\frac\pi2.
$$
For $n\geq1$, <integration by parts> gives
$$
a_n=\frac1\pi\left[\frac{(x+\pi)\sin nx}{n}+\frac{\cos nx}{n^2}\right]_{-\pi}^0
=\frac{1-(-1)^n}{\pi n^2},
$$
and
$$
b_n=\frac1\pi\left[-\frac{(x+\pi)\cos nx}{n}+\frac{\sin nx}{n^2}\right]_{-\pi}^0
=-\frac1n.
$$
Consequently the requested <Fourier series> is
$$
\boxed{\frac\pi4+\sum_{n=1}^{\infty}
\left[\frac{1-(-1)^n}{\pi n^2}\cos nx-\frac1n\sin nx\right].}
$$
It equals the given <function> on each open <continuity> interval. By the <Dirichlet-Jordan convergence theorem>, at $x=2k\pi$ it takes the jump midpoint $\pi/2$, while at $x=(2k+1)\pi$ it takes zero because the periodic one-sided limits agree there.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-ramp-fourier.png]
{title=One period of the ramp-and-zero function, with the Fourier-series value at its jump}
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