= Solution
After discarding $f_0=0$, the preceding <inner products> show that distinct sine functions are <orthogonal>; dividing each by $\sqrt\pi$ makes them <orthonormal>. More precisely, their infinite expansions form an <orthogonal basis> of the odd part of $L^2(-\pi,\pi)$ in the Hilbert-space sense. This is not a claim that every arbitrary odd function is a finite <linear combination> of sines.
Take the <inner product> of $x=\sum_{n\ge1}a_n\sin(nx)$ with $\sin(mx)$. Orthogonality gives
$$
a_m=\frac1\pi\int_{-\pi}^{\pi}x\sin(mx)\,dx=\frac2\pi\left[-\frac{x\cos(mx)}m+\frac{\sin(mx)}{m^2}\right]_0^\pi=\frac{2(-1)^{m+1}}m.
$$
Therefore the <Fourier series of a periodic sawtooth>, scaled to this interval, is
$$
\boxed{x=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n\sin(nx),\qquad-\pi<x<\pi.}
$$
Its periodic extension is piecewise smooth, so the <Dirichlet convergence theorem for Fourier series> gives pointwise convergence to $x$ inside the interval. At the jump endpoints of that periodic extension, the series instead converges to zero, the midpoint of the two one-sided limits.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit-fourier-identity.png]
{title=Fourier sine partial sums approximate the identity inside the interval, while their periodic extension has endpoint jumps}
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