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To determine whether the given series converges or diverges using the Limit Comparison Test, we will follow these steps:

### Series:
\[
\sum_{n=1}^{\infty} \frac{n + 3}{n^3 - 4n + 5}
\]

### Step 1: Identify a comparable series.
We begin by finding a simpler series to compare the given series to. Since the dominant term in the denominator is \(n^3\) (as \(n\) becomes large), we choose the simpler series:
\[
b_n = \frac{1}{n^2}
\]
This is because:
\[
\frac{n + 3}{n^3 - 4n + 5} \sim \frac{n}{n^3} = \frac{1}{n^2} \text{ as } n \to \infty
\]

### Step 2: Apply the Limit Comparison Test.
We compute the limit:
\[
L = \lim_{n \to \infty} \frac{\frac{n + 3}{n^3 - 4n + 5}}{\frac{1}{n^2}}
\]
Simplifying the expression:
\[
L = \lim_{n \to \infty} \frac{(n + 3) n^2}{n^3 - 4n + 5}
\]
As \(n \to \infty\), the highest powers of \(n\) dominate, so we approximate the limit:
\[
L = \lim_{n \to \infty} \frac{n^3 + 3n^2}{n^3 - 4n + 5} = \lim_{n \to \infty} \frac{n^3(1 + \frac{3}{n})}{n^3(1 - \frac{4}{n^2} + \frac{5}{n^3})}
\]
For large \(n\), the terms \(\frac{3}{n}\), \(\frac{4}{n^2}\), and \(\frac{5}{n^3}\) tend to zero, so:
\[
L = \frac{1}{1} = 1
\]

### Step 3: Conclusion based on the Limit Comparison Test.
Since \(L = 1\) and \(0 < L < \infty\), the Limit Comparison Test tells us that the behavior of the given series is the same as that of the simpler series \(\sum \frac{1}{n^2}\). The series \(\sum \frac{1}{n^2}\) is a p-series with \(p = 2 > 1\), which converges.

Thus, the given series:
\[
\sum_{n=1}^{\infty} \frac{n + 3}{n^3 - 4n + 5}
\]
**converges** by the Limit Comparison Test.

---

Would you like more details or have any questions?

Here are five related questions to explore further:
1. How does the Limit Comparison Test differ from the Direct Comparison Test?
2. What are the conditions for using the Limit Comparison Test?
3. Can the Limit Comparison Test be applied if \(L = 0\)?
4. How would the series behave if the denominator's degree was less than 3?
5. What is the convergence behavior of p-series when \(p \leq 1\)?

**Tip**: When using the Limit Comparison Test, always ensure that both series you're comparing are positive and behave similarly as \(n\) approaches infinity.

Use the Limit Comparison Test to determine the convergence or divergence of the series: ∑(n + 3) / (n^3 - 4n + 5) for n=1 to infinity.

Determine whether the series ∑(n + 3) / (n^3 - 4n + 5) converges or diverges using the Limit Comparison Test. This example applies the test by comparing it to a known converging p-series, providing a detailed step-by-step explanation. Suitable for advanced high school or college-level math students.

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