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To determine if the series 
\[
\sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n}
\]
converges or diverges, we can examine the terms and potentially use a comparison test.

### Analysis of the Series Term

The term of the series is:
\[
\frac{\sin^2 n}{2^n}.
\]

1. **Range of \( \sin^2 n \):** Since \(\sin^2 n\) oscillates between 0 and 1 for all integer values of \(n\), we know:
   \[
   0 \leq \sin^2 n \leq 1.
   \]
   
2. **Inequality for the Term**: This implies:
   \[
   0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n}.
   \]
   
### Applying the Comparison Test

The comparison test suggests that if we compare our series with a known convergent or divergent series, we can deduce the convergence or divergence of our original series.

Consider the series:
\[
\sum_{n=1}^{\infty} \frac{1}{2^n}.
\]
This is a geometric series with ratio \( r = \frac{1}{2} \), which is less than 1, and thus converges.

Since 
\[
0 \leq \frac{\sin^2 n}{2^n} \leq \frac{1}{2^n},
\]
and \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) converges, by the **Direct Comparison Test**, we can conclude that:
\[
\sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n}
\]
also converges.

### Conclusion
The correct answer is **A**:
> **A.** Because \(\sum_{n=1}^{\infty} \frac{\sin^2 n}{2^n} \leq \sum_{n=1}^{\infty} \frac{1}{2^n}\) and \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) converges, the series converges by the Direct Comparison Test.

Would you like more details, or do you have any questions?

### Related Questions
1. What is the Direct Comparison Test, and how does it work?
2. How does the behavior of \(\sin^2 n\) affect convergence in this series?
3. What are other methods to test for convergence of series involving trigonometric functions?
4. How would the answer change if the series was \(\sum \frac{\sin n}{2^n}\)?
5. Can we use the Ratio Test for this series, and why or why not?

### Tip
In series comparison problems, finding a known series with similar behavior is a powerful tool to simplify the analysis.

Determine if the series ∑ (sin^2(n) / 2^n) from n=1 to infinity converges or diverges. Use any method, and give a reason for your answer.

Analyze the convergence of the series ∑ (sin^2(n) / 2^n) using the Direct Comparison Test. Understand how trigonometric function behavior and geometric series aid in determining convergence for this series.

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