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To solve this problem, we will follow the steps outlined in the instructions, using the Limit Comparison Test to determine whether the series \( \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{5^n - 4^n}{3^n - 1} \) converges or diverges.

### Step 1: Define the Comparison Series
To apply the Limit Comparison Test, we need a comparison series \( \sum_{n=1}^{\infty} b_n \) that we can use to determine the behavior of \( \sum_{n=1}^{\infty} a_n \). We observe that the terms \( a_n = \frac{5^n - 4^n}{3^n - 1} \) are dominated by the terms \( \frac{5^n}{3^n} \) for large \( n \) (since \( 5^n \) and \( 3^n \) grow much faster than \( 4^n \) and the constant 1). Thus, we choose
\[
b_n = \frac{5^n}{3^n} = \left( \frac{5}{3} \right)^n
\]
as our comparison series.

### Step 2: Express \( b_n \) in the Form \( ar^{n-1} \)
In this case:
- \( a = 1 \)
- \( r = \frac{5}{3} \)

So, \( b_n = \left( \frac{5}{3} \right)^n \) represents a geometric series.

### Step 3: Apply the Limit Comparison Test
The Limit Comparison Test states that we should evaluate the limit:
\[
\lim_{n \to \infty} \frac{a_n}{b_n}
\]
Let's compute this limit:
\[
\frac{a_n}{b_n} = \frac{\frac{5^n - 4^n}{3^n - 1}}{\frac{5^n}{3^n}} = \frac{5^n - 4^n}{3^n - 1} \cdot \frac{3^n}{5^n} = \frac{5^n - 4^n}{5^n} = 1 - \left( \frac{4}{5} \right)^n
\]
Taking the limit as \( n \to \infty \):
\[
\lim_{n \to \infty} \frac{a_n}{b_n} = 1 - 0 = 1
\]

Since this limit is a positive finite number (specifically, 1), the Limit Comparison Test tells us that \( \sum a_n \) and \( \sum b_n \) will either both converge or both diverge.

### Step 4: Determine the Convergence of \( \sum b_n \)
The series \( \sum b_n = \sum \left( \frac{5}{3} \right)^n \) is a geometric series with common ratio \( r = \frac{5}{3} \). Since \( \frac{5}{3} > 1 \), this geometric series **diverges**.

### Conclusion
Since \( \sum b_n \) diverges, \( \sum a_n \) also **diverges** by the Limit Comparison Test.

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### Final Answer
1. **\( a = 1 \)**
2. **\( r = \frac{5}{3} \)**
3. **\( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \)**
4. **\( \sum b_n \) is a divergent series**
5. Therefore, **\( \sum a_n \) diverges by the Limit Comparison Test**

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

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### Related Questions

1. What is the Limit Comparison Test, and how does it work in general?
2. Why did we choose \( \frac{5^n}{3^n} \) as the comparison series?
3. How do we determine whether a geometric series converges or diverges?
4. Can we apply the Limit Comparison Test if the limit is 0 or infinity?
5. What are some other convergence tests that can be used for series?

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### Tip:
When using the Limit Comparison Test, always choose a comparison series that closely resembles the dominant terms of the original series for large \( n \).

Use the Limit Comparison Test to determine whether the series \( \sum_{n=1}^{\infty} \frac{5^n - 4^n}{3^n - 1} \) converges or diverges.

Explore the process of using the Limit Comparison Test to determine the convergence or divergence of the series \( \sum_{n=1}^{\infty} \frac{5^n - 4^n}{3^n - 1} \). This problem involves comparing with a geometric series, evaluating limits, and applying convergence criteria.

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