The series you presented is:

$\sum_{n=1}^{\infty} \sin\left(\frac{8}{n}\right)$

To determine whether this series converges or diverges, let's analyze the behavior of the terms $\sin\left(\frac{8}{n}\right)$ as $n \to \infty$.

1. As $n$ becomes large, $\frac{8}{n}$ becomes very small. For small values of $x$, $\sin(x)$ is approximately $x$, so:

$\sin\left(\frac{8}{n}\right) \approx \frac{8}{n}$

2. This means that for large $n$, the terms of the series behave like $\frac{8}{n}$. The series $\sum_{n=1}^{\infty} \frac{1}{n}$ is known as the harmonic series, which diverges.

Since $\sin\left(\frac{8}{n}\right)$ behaves like $\frac{8}{n}$ for large $n$, and the harmonic series diverges, the series $\sum_{n=1}^{\infty} \sin\left(\frac{8}{n}\right)$ also diverges by comparison to the harmonic series.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the harmonic series and why does it diverge?
2. How can we apply the comparison test to determine convergence?
3. How does the behavior of small-angle approximations help in series analysis?
4. What is the integral test for convergence, and can it be applied here?
5. How does a series involving trigonometric functions typically behave?

Tip: When analyzing the convergence of a series, approximate the terms for large $n$ and compare it to a known series.