# Find convergence or divergence of Series.

a) sum of (sin$(\frac{n \pi}{6}))^n$

b) sum of (sin$(\frac{n \pi}{7}))^n$

I believe that both sequences diverge because a sin or cos function merely loops around the circle periodically and also attains negative values that negate all of the positive values.

I'm unsure about how to proceed with the proofs though. I've tried using the alternating series theorem a few times with no luck.

Ideas/help?

• Hint: Is the limit of the general term equal to $0$? – gniourf_gniourf Apr 16 '14 at 11:33

Denote $$u_n=\sin^n\left(\frac{n\pi}6\right).$$ Then the subsequence $$u_{6n+3}=\sin^{6n+3}\left(\frac{(2n+1)\pi}2\right)=(-1)^{n(6n+3)}$$ doesn't converge to $0$, hence the given series isn't convergent.
Hint: the sequences $\sin (n\pi/6)$ and $\sin (n\pi/7)$ are periodic, so you can view the series as a sum of geometric series. What makes the two different?