Show that $\sum_{k=1}^{\infty} \sin^k (k \pi / 6)$ diverges?

The problem is stated as:

Show that $\sum_{k=1}^{\infty} \sin^k (k \pi / 6)$ diverges?

My attempt:

First of all, I thought that for $k = 6n$ for some natural number $n$, we have that the term is equal to $0$. Then, for $k = 6n-3$, we have that the general term is equal to $(-1)^{n-1}$ and so forth. However, this quickly becomes hard since I have to split the sum into all of these possible outcomes for the term.

I then thought since the summation can be split into multiple ones, where one of the summand contains $(-1)^{n-1}$ which clearly diverges, then the whole sum must diverge. However, in a later problem, I have to show that $\sum_{k=1}^{\infty} \sin^k (k \pi / 7)$ converges instead. However, since k is discrete, there doesn't exist some $k's$ such that we can get $(-1)^{n-1}$ as in our previous case. Maybe this shows that I'm on the right path, but it can't be enough to prove divergence I think?

I wonder what method I can apply here. The general term doesn't tend to $0$ as k tends to infinity, and maybe this argument is enough to prove that it diverges?

Thank you for any help.

Solution 1:

Let $a_k=\sin^k(k\pi/6)$.

As you noticed, $a_{12k}=1$ so $\lim_{k\to\infty}a_k\neq 0$, which implies that the series $\sum_{k=1}^\infty a_k$ is divergent (by the Divergence Test).

Now, let $b_k=\sin(k\pi/7)^k$ and $c_k=\sin(k\pi/7)$. The sequence $c_k$ is $14$-periodic, so $c=Max\{|c_k|, k\ge 1\}$ exists. Clearly, $c<1$. We have $c_k\le c<1$ so

$$|b_k|\le c^k$$

Since $|c|<1$, $\sum_{k=1}^\infty c^k$ is convergent, therefore $\sum_{k=1}^\infty b_k$ is absolutely convergent, hence convergent.