 Solution

Since $|T_j(x)|\le 1$ on $[-1,1]$ and the coefficients are summable, the Weierstrass M-test gives uniform convergence of this positive lacunary Chebyshev series and makes its sum continuous.

Let $r$ be the largest integer with $3^r\le n$, and set $r=-1$ when $n=0$. The candidate and its tail size areThe empty sum for $n=0$ is zero. The candidate has degree at most $n$. The triangle inequality bounds its error by the proposed tail sum, and equality holds at $x=1$, since every Chebyshev polynomial has $T_j(1)=1$ and all tail coefficients are positive. It remains to prove that another polynomial cannot improve this error.

Put $L=3^{r+1}$, the first omitted degree, so $L>n$. At the points $y_j=\cos (j\pi /L)$ for $j=0,\dots ,L$, every omitted term hasbecause $3^{k-r-1}$ is odd. Uniform convergence therefore givesThese $L+1$ distinct points run in decreasing order; reversing their order still gives alternating signs. Since $L+1\ge n+2$, select $n+2$ consecutive ones and apply the Chebyshev alternation theorem. It proves both optimality and uniqueness of the displayed candidate. Thus the best polynomial and the error stay unchanged between successive powers of three; this exact truncation property depends on the synchronized tail signs, not on a general rule that truncating a Chebyshev expansion is optimal.