= 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 satisfies
$$
r(x):=f(x)-t_n(x)=\sum_{k=m+1}^\infty a_k\cos(5^kx),\qquad \|r\|_\infty\le A.
$$
At $x_j=j\pi/q$, for $j=0,\ldots,2q-1$, the <integer> $5^k/q$ is odd for every omitted frequency. Thus $\cos(5^kx_j)=(-1)^j$ and $r(x_j)=(-1)^jA$. In particular $\|r\|_\infty=A$.
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 \b[the proposed partial sum is the unique best approximant], with
$$
\boxed{E_n(f)=\sum_{k=m+1}^\infty a_k\qquad(5^m\le n<5^{m+1}).}
$$
This is an instance of <positive lacunary trigonometric series>: all omitted odd frequency ratios align their extrema.
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