= Solution
<Termwise differentiation of a Fourier series>, applied to the continuous periodic extension in part (c), gives for $-1<x<1$
$$
2x=-\frac4\pi\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin(n\pi x).
$$
Hence the required <Fourier sine series> is
$$
\boxed{x=\frac2\pi\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n\sin(n\pi x),\qquad-1<x<1.}
$$
The endpoint deserves care: at $x=1$ every term is zero, so the <Fourier series> sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one <point> does not change any <Fourier coefficient>, so this is still the <Fourier series> of the stated <function> on $(-1,1]$.
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