Consider an ideal low pass filter $H(\omega)$, and the input to this filter is the periodic square wave $x(t)$. Find the output $y(t)$
So you actually solve(d) this problem by computing the CTFS of the input $x(t)$: 
$$ C_k = 5 \frac{ 1 - e^{ - j k \pi } }{ j k \pi } = \begin{cases} 0 &amp;, \text{ for k even , } k \neq 0 \\ 10/jk\pi &amp;, \text{ for k odd} \end{cases} $$ which gives the coefficient associated with the complex exponential of frequency $\omega_k = k \omega_0$ for the fundamental frequency $\omega_0 = \frac{ 2 \pi }{T_0} = \frac{ 2 \pi }{2}  = \pi$ , as the period of the input $x(t)$ is $2$ seconds.

Since the ideal lowpass filter passes those frequencies below its cutoff frequency $\omega_c = 4 \pi$, then we would only take those coefficiencs whose frequencies fall below $\omega_c$ yielding those values of $k = 0,\pm 1, \pm 3$ . Those indices higher than $\pm 5$ are filtered out by the ideal lowpass filter.

Therefore the output $y(t)$ is: 
$$ y(t) = 5 + \frac{10}{j\pi} \left( e^{j\pi t} - e^{-j\pi t} \right) + \frac{10}{j3\pi} \left( e^{j3\pi t} - e^{-j3\pi t} \right) $$ which simplifies to: 
$$ y(t) = 5 + \frac{20}{\pi} \sin(\pi t) + \frac{20}{3\pi} \sin(3\pi t) $$