If, in the periods, the two half's signal periodic have the same form and opposite phases, the periodic signal has symmetry of half wave. If the periodic signal $g(t)$, of period $T_0$, satisfy the condition of symmetry of half wave, so:
$g(t-\frac{T_0}{2}) = -g(t)$
In this case, proof that all the harmonics (coefficient) of even order are NULL.
Let the function $g(t)$ has Fourier series coefficient of $a_k$ for the harmonic $e^{i2\pi k/T_0}$, $k$ are integers from $-\infty$ to $+\infty$.
$g\left(t-\frac{T_0}{2}\right)$ is a time-shifted function of $g(t)$, and has corresponding Fourier series coefficients of $a_k e^{-i\pi k}$ for the harmonic $e^{i2\pi k/T_0}$.
Also, $-g(t)$ has Fourier series coefficients of $-a_k$.
From the given condition, the Fourier series coefficients of both $g\left(t-\frac{T_0}{2}\right)$ and $-g(t)$ should be equal for all $k$, i.e.
$$\begin{align} a_k e^{-i\pi k} =& -a_k\\ a_k \left(e^{-i\pi k} + 1\right) =& 0 \end{align}$$
For even $k$s, the second factor is non-zero, therefore $a_k$ must be zero. In other words, there are no even order harmonics. (For odd $k$s, the second factor is zero, and $a_k$ may or may not be zero)