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To solve this problem using the Limit Comparison Test, let's go through it step-by-step.

### Step 1: Identify the series \( a_n \)

The series given is:
\[
\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{9n^2 + 9n - 1}{(n+2)^4}.
\]

### Step 2: Choose a comparison series \( b_n \)

To apply the Limit Comparison Test, we need to find a suitable comparison series \( b_n \). We should choose \( b_n \) to be a simpler series that has a similar growth rate to \( a_n \) as \( n \to \infty \).

Notice that the highest power term in the numerator of \( a_n \) is \( 9n^2 \) and in the denominator is \( (n+2)^4 \), which approximates to \( n^4 \) as \( n \) becomes large. Thus, a good choice for \( b_n \) would be:
\[
b_n = \frac{1}{n^2}.
\]

This choice simplifies the comparison because \( \frac{1}{n^2} \) is a known \( p \)-series with \( p = 2 \).

### Step 3: Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} \)

Now, we calculate the limit:
\[
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{9n^2 + 9n - 1}{(n+2)^4}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{9n^2 + 9n - 1}{(n+2)^4} \cdot n^2.
\]

Simplifying further:
\[
= \lim_{n \to \infty} \frac{9n^4 + 9n^3 - n^2}{(n+2)^4}.
\]

As \( n \to \infty \), the highest power terms dominate, so we approximate by focusing on \( \frac{9n^4}{n^4} = 9 \). Therefore:
\[
\lim_{n \to \infty} \frac{a_n}{b_n} = 9.
\]

Since this limit is a positive finite number, the Limit Comparison Test applies.

### Step 4: Determine convergence of \( b_n \)

The comparison series \( \sum b_n = \sum \frac{1}{n^2} \) is a \( p \)-series with \( p = 2 \), which is known to converge (since \( p > 1 \)).

### Conclusion

Since \( \sum b_n \) converges and \( \lim_{n \to \infty} \frac{a_n}{b_n} \) is finite and positive, the original series \( \sum a_n \) also converges by the Limit Comparison Test.

---

### Answer

1. **Comparison series** \( b_n = \frac{1}{n^2} \).
2. **Value of \( p \)** in the comparison series: \( p = 2 \).
3. **Limit** \( \lim_{n \to \infty} \frac{a_n}{b_n} = 9 \).
4. \( \sum b_n \) is a **convergent \( p \)-series**.
5. Therefore, \( \sum a_n \) **converges** by the Limit Comparison Test.

Would you like more details on any of these steps or have additional questions?

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### Related Questions

1. How do you choose a comparison series for the Limit Comparison Test?
2. What are the conditions required for a \( p \)-series to converge?
3. Can the Limit Comparison Test be applied if the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \) is zero or infinity?
4. What is the difference between the Limit Comparison Test and the Direct Comparison Test?
5. How do you determine the dominant terms in a series to select an appropriate \( b_n \)?

### Tip

For the Limit Comparison Test, always focus on the highest power terms in the numerator and denominator of \( a_n \) to find an appropriate \( b_n \).

Use the Limit Comparison Test to determine whether the series \( \sum_{n=1}^{\infty} \frac{9n^2 + 9n - 1}{(n+2)^4} \) converges or diverges.

Explore the use of the Limit Comparison Test to determine the convergence of the series \( \sum \frac{9n^2 + 9n - 1}{(n+2)^4} \). This solution includes steps for choosing a comparison series, calculating limits, and analyzing convergence using p-series criteria. Suitable for college-level calculus students.

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