The reason is simple: when a> is at 1, the equivalent infinitesimal n→∞, sin (π/3n) ~ (π/3n).

Then lim [n→∞] (a n) sin (π/3 n)

= lim[n→∞] (a^n)(π/3^n)

= lim[n→∞] π(a/3)^n

= π * lim [n→∞] (a/3) n = 0, just a/3.

In addition, when a= 1, lim [n→∞] (an) sin (π/3n) = lim [n→∞] sin (π/3n) = 0 holds.

So there is a= 1, which still converges and does not diverge.

Of course, therefore, 0