= Solution
For positive integers $n$, the summand is $a_n=\sin(\pi\sin(1/n))/n$. Since $0<\sin(1/n)<1$, these terms are positive. Apply the <limit comparison test> with the convergent <p-series> $\sum n^{-2}$:
$$
\frac{a_n}{n^{-2}}
=n\sin(\pi\sin(1/n))
=\frac{\sin(\pi\sin(1/n))}{\pi\sin(1/n)}\,
\pi\frac{\sin(1/n)}{1/n}
\longrightarrow\pi.
$$
Each sine ratio tends to one. The <limit comparison test> says that two positive-term <series> whose term ratio tends to a finite positive number are either both <convergent series>[convergent] or both <divergent series>[divergent]. Therefore
$$
\boxed{\sum_{n=1}^\infty f_{-1}(n)\text{ converges absolutely}}.
$$
There is also a direct <comparison test for series>: $0\leq\sin(\pi\sin(1/n))\leq\pi\sin(1/n)\leq\pi/n$, giving $0\leq a_n\leq\pi/n^2$.
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