Solution: When ∵n→∞, sin (π/3 n) ~ π/3 n, ∴ series ∑ (an n) sin (π/3 n) and series ∑ (an n) (π/3 n) = π ∑ (a).

∴∴ (a n) sin (π/3 n) convergence, ∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴.

Similarly, a>0, ∴ 0 < a & lt3。

For reference.