How did we get the last equation? Why can the summation be converted into a square term?
$$\begin{align}K_m(x)&:=1+\frac{2}{m}\sum_{j=1}^{m-1}(m-j)\cos(jx)\\&= \frac1m\sum_{j=-(m-1)}^{m-1} (m - |j|)e^{ijx}\\ &= \frac{1}{m} \sum_{j=0}^{2(m-1)} (m - |j - (m - 1)|)e^{i(j-(m-1))x} \\ &= \frac{1}{m} \left(\sum_{k=0}^{m-1} e^{i\left(k - \frac{m-1}{2}\right)x}\right)^2 \end{align}$$