= Solution
The <Chebyshev polynomial> identity $T_m(\cos\theta)=\cos(m\theta)$ implies $|T_m(x)|\leq1$ on $[-1,1]$. Since $\sum_i\gamma_i<\infty$, the <Weierstrass M-test> shows that
$$
f(x)=\sum_{i=1}^{\infty}\gamma_iT_{3^i}(x)
$$
converges uniformly on $[-1,1]$. Every partial sum is continuous, so the <uniform limit theorem> makes $f$ a well-defined continuous function.
Let
$$
P_n=\sum_{i=1}^n\gamma_iT_{3^i},
\qquad
x_k=-\cos\frac{k\pi}{3^{n+1}}
\quad(0\leq k\leq3^{n+1}).
$$
Then $-1=x_0<x_1<\cdots<x_{3^{n+1}}=1$. Since every $3^i$ is odd and, for $i\geq n+1$, $3^{i-n-1}$ is an odd integer,
$$
T_{3^i}(x_k)
=-\cos\left(k\pi3^{i-n-1}\right)=(-1)^{k+1}.
$$
It follows that
$$
\boxed{f(x_k)-P_n(x_k)
=(-1)^{k+1}\sum_{i=n+1}^{\infty}\gamma_i}.
$$
Thus the error has equal alternating extrema at $3^{n+1}+1$ ordered points. The <Chebyshev alternation theorem> shows that $P_n$ is a best uniform approximation among polynomials of degree at most $3^{n+1}-1$, and in particular
$$
E_{3^{n+1}-1}(f)
:=\inf_{\deg P\leq3^{n+1}-1}\lVert f-P\rVert_\infty
\geq\sum_{i=n+1}^{\infty}\gamma_i.
$$
Now let $(\delta_N)$ be the prescribed decreasing positive sequence. Define
$$
s_0=\delta_1+1,
\qquad
s_j=\delta_{3^j}+2^{-j}\quad(j\geq1),
\qquad
\gamma_j=s_{j-1}-s_j.
$$
Then $s_j$ decreases strictly to zero, so every $\gamma_j>0$, the series $\sum_j\gamma_j=s_0$ converges, and
$$
\sum_{i=j+1}^{\infty}\gamma_i=s_j\geq\delta_{3^j}.
$$
Given $N\geq1$, choose $j\geq0$ with $3^j\leq N\leq3^{j+1}-1$. Monotonicity of best-approximation errors and of $\delta_N$ gives
$$
E_N(f)\geq E_{3^{j+1}-1}(f)
\geq s_j
\geq\delta_{3^j}
\geq\delta_N.
$$
Hence this continuous $f$ satisfies
$$
\boxed{\sup_{t\in[-1,1]}|f(t)-P(t)|\geq\delta_N}
$$
for every polynomial $P$ of degree at most $N$. This is the polynomial form of <Bernstein's lethargy theorem>.
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