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.

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?

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?

Tip:

When using the Limit Comparison Test, always choose a comparison series that closely resembles the dominant terms of the original series for large $n$.