 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 $\sum n^{-2}$: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 or both divergent. ThereforeThere is also a direct comparison test for series: $0\le \sin (\pi \sin (1/n))\le \pi \sin (1/n)\le \pi /n$, giving $0\le a_n\le \pi /n^2$.