Because the forces have been applied for a long time, any non-zero damping eliminates the homogenous parts of the solution. We need then only consider the particular solution for each \( n \).
The problem is considered solved if one can find expressions for the coefficients \( f_n \) and \( g_n \), even though the solutions are expressed as an infinite sum. The coefficients can be extracted from the function \( F(t) \) by
$$ \begin{eqnarray} \tag{4} f_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\cos(2n\pi t/\tau),\\ \nonumber g_n&=&\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~F(t)\sin(2n\pi t/\tau). \end{eqnarray} $$To check the consistency of these expressions and to verify Eq. (4), one can insert the expansion of \( F(t) \) in Eq. (3) into the expression for the coefficients in Eq. (4) and see whether
$$ f_n=\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~\left\{\frac{f_0}{2}+\sum_{m>0}f_m\cos(m\omega t)+g_m\sin(m\omega t)\right\}\cos(n\omega t). $$Immediately, one can throw away all the terms with \( g_m \) because they convolute an even and an odd function. The term with \( f_0/2 \) disappears because \( \cos(n\omega t) \) is equally positive and negative over the interval and will integrate to zero. For all the terms \( f_m\cos(m\omega t) \) appearing in the sum, one can use angle addition formulas to see that \( \cos(m\omega t)\cos(n\omega t)=(1/2)(\cos[(m+n)\omega t]+\cos[(m-n)\omega t] \). This will integrate to zero unless \( m=n \). In that case the \( m=n \) term gives
$$ \begin{equation} \int_{-\tau/2}^{\tau/2}dt~\cos^2(m\omega t)=\frac{\tau}{2}, \tag{5} \end{equation} $$and
$$ f_n=\frac{2}{\tau}\int_{-\tau/2}^{\tau/2} dt~f_n/2=f_n. $$The same method can be used to check for the consistency of \( g_n \).