Series Question involving sines?

Use the geometric series

1/(1 - t) = Σ(n=0 to ∞) tⁿ.

Letting x = e^(ix)/3 (valid, since |e^(ix)/3| = 1/3 < 1):

1/(1 - e^(ix)/3) = Σ(n=0 to ∞) (e^(ix)/3)ⁿ

==> 3/(3 - e^(ix)) = Σ(n=0 to ∞) e^(nix) / 3ⁿ.

However, e^(nix) = cos(nx) + i sin(nx) by Euler's identity. So,

3/[3 - (cos x + i sin x)] = Σ(n=0 to ∞) [cos(nx) + i sin(nx)] / 3ⁿ.

The left side may be rewritten

3/[(3 - cos x) - i sin x]

= 3[(3 - cos x) + i sin x] / {[(3 - cos x) - i sin x] [(3 - cos x) + i sin x]}, by conjugates

= [(9 - 3 cos x) + i (3 sin x)] / [(3 - cos x)² + sin² x]

= [(9 - 3 cos x) + i (3 sin x)] / (10 - 6 cos x).

Therefore,

[(9 - 3 cos x) + i (3 sin x)] / (10 - 6 cos x) = Σ(n=0 to ∞) [cos(nx)/3^n + i sin(nx)/ 3ⁿ].

Equating the imaginary parts yields

Σ(n=1 to ∞) sin(nx)/3ⁿ = 3 sin x / (10 - 6 cos x).

Note that the n = 0 term contributes nothing to the series.

Since sin² x + cos² x = 1, we have sin x = 1/5 ==> cos x = √(24)/5 = (2/5)√6.

**Note: Here, I assumed that x is in the first quadrant. If x is not in the first quadrant, use cos x = (-2/5)√6.

Therefore,

Σ(n=1 to ∞) sin(nx)/3ⁿ = 3(1/5) / [10 - 6 (2/5)√6] = 3 / (50 - 12 √6).

I hope this helps!