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To determine the convergence or divergence of each series in the problem, we'll analyze each part using convergence tests.

### Problem Statement
Determine whether the following series are convergent or divergent:

1. \(\sum_{n=1}^{\infty} \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}}\)
2. \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6}\)

Let's proceed with each series separately.

---

### Part (a): \(\sum_{n=1}^{\infty} \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}}\)

To analyze this series, let's examine the general term:
\[
a_n = \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}}
\]

#### Step 1: Simplify the Dominant Terms
For large \( n \), the terms \(3n^2\) in the numerator and \(2n^3\) in the denominator dominate the behavior of \(a_n\). So let's approximate:
\[
a_n \approx \frac{\sqrt[3]{3n^2}}{\sqrt[4]{2n^3}} = \frac{(3n^2)^{1/3}}{(2n^3)^{1/4}}
\]

Simplifying each part:
\[
(3n^2)^{1/3} = 3^{1/3} n^{2/3} \quad \text{and} \quad (2n^3)^{1/4} = 2^{1/4} n^{3/4}
\]

Thus,
\[
a_n \approx \frac{3^{1/3} n^{2/3}}{2^{1/4} n^{3/4}} = \frac{3^{1/3}}{2^{1/4}} n^{-1/12}
\]

#### Step 2: Determine the Convergence of \(\sum_{n=1}^{\infty} n^{-1/12}\)
The term \(a_n\) behaves like \(n^{-1/12}\) for large \(n\). To determine the convergence of this series, we can compare it to the \(p\)-series \(\sum n^{-p}\).

Since \(p = \frac{1}{12}\), which is less than 1, the \(p\)-series diverges. Therefore, the original series also diverges by the Comparison Test.

**Conclusion for (a):** The series \(\sum_{n=1}^{\infty} \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}}\) **diverges**.

---

### Part (b): \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6}\)

In this case, we have an alternating series due to the factor \((-1)^{n+1}\).

#### Step 1: Consider the Alternating Series Test (Leibniz's Test)
For the alternating series test, we need to check:
1. \( b_n = \left|\frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6}\right| = \frac{n^2}{4n^2 + 5n + 6} \) is decreasing.
2. \(\lim_{n \to \infty} b_n = 0\).

#### Step 2: Check if \(b_n\) is Decreasing
The term \(b_n = \frac{n^2}{4n^2 + 5n + 6}\) simplifies by observing that, for large \(n\), the dominant terms in both the numerator and the denominator are \(n^2\) and \(4n^2\), respectively:
\[
b_n \approx \frac{n^2}{4n^2} = \frac{1}{4}
\]

Since \(b_n\) does not approach zero (it approaches \(\frac{1}{4}\) instead), the Alternating Series Test fails.

#### Step 3: Use the Limit Comparison Test
Since \(b_n \approx \frac{1}{4}\) for large \(n\), the series does not decay to zero, meaning that \(\sum_{n=1}^{\infty} b_n\) diverges. Hence, the original series also diverges.

**Conclusion for (b):** The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6}\) **diverges**.

---

### Final Answer
Both series **diverge**.

Would you like further clarification on any step? 

---

#### Related Questions:
1. How would the behavior change if the exponent in \(n\) were different in part (a)?
2. What would happen if part (b) did not have the alternating factor \((-1)^{n+1}\)?
3. Can the Limit Comparison Test be applied in part (a) directly with a \(p\)-series?
4. How would we handle a similar problem if higher powers of \(n\) were present in the denominator?
5. What is the general \(p\)-series test, and how can it be applied here?

#### Tip:
When faced with series that have alternating terms, the Alternating Series Test is often helpful, but make sure the terms decrease to zero.

Determine whether the series \( \sum_{n=1}^{\infty} \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}} \) and \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6} \) are convergent or divergent.

Analyze the convergence or divergence of the series \( \sum_{n=1}^{\infty} \frac{\sqrt[3]{3n^2 + 1}}{\sqrt[4]{2n^3 + 3n + 5}} \) and \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{4n^2 + 5n + 6} \) using convergence tests like the Comparison Test, p-Series Test, and Alternating Series Test.

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