Prove that $\sum_{n\in\mathbb{N}}\sin^3{\frac{2}{n}}$ converges

Question

$$\sum_{n\in\mathbb{N}}\sin^3{\frac{2}{n}}$$

The only thing I can think of is that sinus cubed behaves similarly to the $\frac{(-1)^{n + 1}}{n}$ series, but that really leads me nowhere... I am completely stuck.

How would you go about showing this series converges? Thank you.

Answer 1

Since $(\forall x\in\mathbb{R}):\bigl|\sin x\bigr|\leqslant|x|$, $(\forall n\in\mathbb{N}):\left|\sin^3\frac2n\right|\leqslant\frac8{n^3}$. Now, apply the comparison test.

Answer 2

Note that $\sin(x) \sim x$, as $x \to 0$, so we have that $\sin^3\left( \frac 2n \right) \sim \left( \frac{2}{n} \right)^3$ as $ n \to \infty$ So use the limit comparison test to show that the series converges absolutely.

Answer 3

You can use the inequalities $0\le\sin x\le x$ for $0\le x\le\pi$ and $0\le{2\over n}\le\pi$ for $n\ge1$ to conclude

$$0\le\sum\sin^3\left(2\over n\right)\le\sum\left(2\over n\right)^3=8\sum{1\over n^3}$$