= Solution
The <geometric series> formula and part (b) give
$$
E_n(f)=\sum_{k=m+1}^\infty3^{-k}=\frac12\,3^{-m},\qquad 5^m\le n<5^{m+1}.
$$
Put $\alpha=\log 3/\log 5$. Then $5^\alpha=3$, so $n^\alpha<3^{m+1}$ and
$$
\boxed{E_n(f)\le\frac{3}{2}n^{-\log 3/\log 5}.}
$$
This is the \b[largest possible decay exponent]: at $n=5^m$, an estimate with exponent $\beta>\alpha$ would require $\frac12\le c(3/5^\beta)^m$, whose right side tends to zero. Any smaller exponent also gives a valid weaker estimate. In fact $\frac12n^{-\alpha}\le E_n(f)<\frac32n^{-\alpha}$ for every $n\ge1$.
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