 Solution

The Weierstrass M-test gives uniform convergence and continuity of the series. For the proposed trigonometric polynomial, set $q=5^{m+1}$ and $A={\sum }_{k=m+1}^{\infty }{a}_k$. The tail satisfiesAt $x_j=j\pi /q$, for $j=0,\dots ,2q-1$, the integer $5^k/q$ is odd for every omitted frequency. Thus $\cos (5^k{x}_j)=(-1{)}^j$ and $r(x_j)=(-1{)}^{j}A$. In particular $\Vert r{\Vert }_{\infty }=A$.

Since $n<q$, these $2q$ points include at least $2n+2$ consecutive alternating extrema in a single period. Also $5^m\le n$, so $t_n\in {\mathcal{T}}_n$. The trigonometric Chebyshev alternation theorem proves that the proposed partial sum is the unique best approximant, withThis is an instance of positive lacunary trigonometric series: all omitted odd frequency ratios align their extrema.