is it possible to solve sigma (n=1 and infinity) sin(8/n)

The series in question is:

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

To determine if this series converges or diverges, we can use a few tools from calculus and analysis. Let’s explore the main points:

### 1. Behavior of the Terms:
- As \( n \to \infty \), \(\sin\left(\frac{8}{n}\right)\) behaves like \(\frac{8}{n}\), since for small angles (in radians), \(\sin(x) \approx x\).
  
  Thus, we can approximate the series as:
  \[
  \sum_{n=1}^{\infty} \sin\left(\frac{8}{n}\right) \approx \sum_{n=1}^{\infty} \frac{8}{n}.
  \]

### 2. Compare to a Known Series (Harmonic Series):
- The series \(\sum_{n=1}^{\infty} \frac{8}{n}\) is a constant multiple of the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\).
- The harmonic series is known to **diverge**, which means that it grows without bound as \( n \) increases.

### 3. Applying the Comparison Test:
- Since \(\sin\left(\frac{8}{n}\right)\) is positive for all \( n \geq 1 \) and behaves similarly to \(\frac{8}{n}\), we can use the **Comparison Test**.
- If \(\sum a_n\) diverges and \( 0 \leq b_n \leq a_n \) for all \( n \geq N \) for some positive integer \( N \), then \(\sum b_n\) also diverges.
  
  Here, \( a_n = \frac{8}{n} \) and \( b_n = \sin\left(\frac{8}{n}\right) \), so since \(\sum \frac{8}{n}\) diverges, the original series also **diverges**.

### Conclusion:
The series \(\sum_{n=1}^{\infty} \sin\left(\frac{8}{n}\right)\) **diverges**.

Would you like more details or have any questions about this solution? Here are a few related questions to expand your understanding:

1. How does the harmonic series \(\sum \frac{1}{n}\) behave, and why is it divergent?
2. What is the Comparison Test, and how is it used to determine the convergence of a series?
3. How does \(\sin(x)\) behave for very small values of \( x \)?
4. What other methods can be used to determine the convergence of a series?
5. How does the concept of asymptotic behavior help in analyzing series?

**Tip:** For small angles \( x \), \(\sin(x) \approx x\) is a useful approximation. This is often used in series analysis to simplify trigonometric terms.

Is it possible to solve sigma (n=1 and infinity) sin(8/n)?

Explore the convergence of the series Σ sin(8/n) from n=1 to infinity using calculus techniques such as the Comparison Test. This solution includes an analysis of the harmonic series and the small-angle approximation for trigonometric functions, suitable for students in Grades 11-12 and college-level calculus.

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