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The series given in the problem is:

\[
\sum_{n=1}^{\infty} \frac{\sin^2(n)}{2^n}
\]

We are asked to determine whether this series converges or diverges, using the Direct Comparison Test.

### Direct Comparison Test:
The Direct Comparison Test compares the terms of one series to another series whose convergence behavior is known. If we can show that the terms of our series are bounded by the terms of a convergent series, then our series also converges. Conversely, if the terms are greater than or equal to those of a divergent series, then the series diverges.

### Step 1: Analyze the Behavior of \( \sin^2(n) \)
The function \( \sin^2(n) \) is bounded between 0 and 1 for all values of \( n \). Therefore, we have:

\[
0 \leq \sin^2(n) \leq 1 \quad \text{for all} \ n
\]

Thus, the terms of the series satisfy:

\[
0 \leq \frac{\sin^2(n)}{2^n} \leq \frac{1}{2^n}
\]

### Step 2: Comparison with a Known Series
We now compare the given series to the geometric series:

\[
\sum_{n=1}^{\infty} \frac{1}{2^n}
\]

This is a geometric series with a ratio \( r = \frac{1}{2} \), which is less than 1. Therefore, the series:

\[
\sum_{n=1}^{\infty} \frac{1}{2^n}
\]

**converges**.

Since:

\[
0 \leq \frac{\sin^2(n)}{2^n} \leq \frac{1}{2^n}
\]

and \( \sum \frac{1}{2^n} \) converges, by the Direct Comparison Test, the given series:

\[
\sum_{n=1}^{\infty} \frac{\sin^2(n)}{2^n}
\]

**also converges**.

### Conclusion:
The series converges, and the correct answer is **A**.

Let me know if you'd like further details or have any other questions!

Here are some related questions:
1. What is the Direct Comparison Test and how is it applied?
2. How can we compare a series with bounded functions like \( \sin^2(n) \)?
3. Why does the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converge?
4. What is the difference between the Direct Comparison Test and the Limit Comparison Test?
5. Can the series \( \sum_{n=1}^{\infty} \frac{\sin^2(n)}{n^2} \) be tested for convergence in a similar way?

**Tip:** When using the Direct Comparison Test, always ensure that the series you are comparing with is a well-known series (like a p-series or geometric series).

Determine if the series converges or diverges: \( \sum_{n=1}^{\infty} \frac{\sin^2(n)}{2^n} \). Use any method, and give a reason for your answer.

Learn how to determine whether the series \( \sum_{n=1}^{\infty} \frac{\sin^2(n)}{2^n} \) converges or diverges using the Direct Comparison Test. This problem involves comparing to a known convergent geometric series, and is suitable for college-level calculus students.

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