The Taylor series expansion of $\sin x$ around $0$ is given by:
$$
\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
$$
So, we can write $\sin \frac{1}{n}$ as:
$$
\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{3!n^3} + \frac{1}{5!n^5} -
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