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.

Understand the divergence of the infinite series ∑ sin(8/n) by applying the harmonic series comparison and the small-angle approximation for trigonometric functions. This detailed solution is ideal for advanced high school or undergraduate students studying series and convergence tests.

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