= Solution
An explicit construction gives <Bernstein's lethargy theorem> in the requested inequality form. Set
$$
a_0=\epsilon_0-\epsilon_1,\qquad
a_k=\epsilon_{3^{k-1}}-\epsilon_{3^k}\quad(k\ge1).
$$
Strict decrease makes every coefficient positive. Telescoping and the limit assumption give
$$
\sum_{k=0}^\infty a_k=\epsilon_0,\qquad
\sum_{k=K}^\infty a_k=\epsilon_{3^{K-1}}\quad(K\ge1).
$$
Thus
$$
f_\epsilon(x)=\sum_{k=0}^\infty a_kT_{3^k}(x)
$$
defines a <continuous function> by the <Weierstrass M-test> and the <uniform limit theorem>. Apply the <positive lacunary Chebyshev series> calculation. At $n=0$,
$$
E_0(f_\epsilon)=\epsilon_0.
$$
For $n\ge1$, choose $K\ge1$ so that $3^{K-1}\le n<3^K$. Its error is
$$
E_n(f_\epsilon)=\epsilon_{3^{K-1}}\ge\epsilon_n.
$$
Therefore the function satisfies
$$
\boxed{E_n(f_\epsilon)\ge\epsilon_n\quad\text{for every }n\ge0}.
$$
This <explicit Chebyshev construction for Bernstein lethargy> also has $E_n(f_\epsilon)\to0$. It shows that continuity imposes no universal speed of convergence of best <polynomial> approximation, even though convergence itself follows from the <Weierstrass approximation theorem>.
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