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.