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Let's examine each series provided and determine whether they are convergent or divergent using appropriate tests.

### a) \( \sum_{n=1}^{\infty} \frac{n^2 - 1}{2n^2 + 3} \)

#### Test: **Limit Comparison Test**
We will compare this series to a simpler known series, particularly \( \frac{n^2}{2n^2} = \frac{1}{2} \), which is similar for large \( n \).

1. Compute the limit of the ratio of the given series' terms to the comparison series:
   \[
   \lim_{n \to \infty} \frac{\frac{n^2 - 1}{2n^2 + 3}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(n^2 - 1) n^2}{(2n^2 + 3)} = \frac{1}{2}.
   \]
   This limit is a finite non-zero number.

Since the series \( \sum \frac{1}{n^2} \) is a **convergent p-series** (with \( p = 2 \)), by the limit comparison test, the given series also **converges**.

### b) \( \sum_{n=2}^{\infty} \frac{3^n n^2}{n!} \)

#### Test: **Ratio Test**
We'll use the ratio test to analyze the behavior of this series.

1. Let \( a_n = \frac{3^n n^2}{n!} \).
2. Compute the ratio of successive terms:
   \[
   \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{3^{n+1} (n+1)^2}{(n+1)!} \cdot \frac{n!}{3^n n^2} = \lim_{n \to \infty} \frac{3(n+1)^2}{(n+1)n^2}.
   \]
   Simplifying:
   \[
   = \lim_{n \to \infty} \frac{3(n^2 + 2n + 1)}{n^3 + n^2} \approx \frac{3}{n}.
   \]
   As \( n \to \infty \), the limit tends to \( 0 \).

Since the limit is less than 1, the series **converges** by the **Ratio Test**.

### c) \( \sum_{n=2}^{\infty} \frac{3^n + 2^n}{4^n} \)

#### Test: **Comparison Test**
Let's break this series into two simpler ones:
\[
\sum_{n=2}^{\infty} \frac{3^n}{4^n} + \sum_{n=2}^{\infty} \frac{2^n}{4^n}.
\]
Both parts are geometric series.

1. The first term can be written as:
   \[
   \sum_{n=2}^{\infty} \left( \frac{3}{4} \right)^n.
   \]
   Since \( \frac{3}{4} < 1 \), this geometric series converges.

2. The second term:
   \[
   \sum_{n=2}^{\infty} \left( \frac{1}{2} \right)^n.
   \]
   This is also a geometric series with ratio \( \frac{1}{2} \), so it converges as well.

Since both geometric series converge, the original series also **converges** by the **Comparison Test**.

---

### Summary:
- **a)** Converges by **Limit Comparison Test** with \( \frac{1}{n^2} \).
- **b)** Converges by **Ratio Test**.
- **c)** Converges by **Comparison Test** with geometric series.

Would you like more detailed steps or further explanation?

### Related Questions:
1. What is the Limit Comparison Test, and when is it used?
2. How do you apply the Ratio Test in series convergence?
3. Can the Comparison Test be used for any type of series?
4. What are p-series, and how do they help in determining convergence?
5. What are the criteria for a geometric series to converge?

### Tip:
Always ensure you compare or test series with well-known simpler ones (like geometric or p-series) when possible to simplify the analysis of convergence.

Determine which of the following series are convergent and which are divergent. You must provide sufficient justification for your answer. Clearly state the test/method used. a) \(\sum_{n=1}^{\infty} \frac{n^2 - 1}{2n^2 + 3}\), b) \(\sum_{n=2}^{\infty} \frac{3^n n^2}{n!}\), c) \(\sum_{n=2}^{\infty} \frac{3^n + 2^n}{4^n}\)

This problem explores the convergence of different infinite series using key methods such as the Limit Comparison Test, Ratio Test, and Comparison with geometric series. Learn to analyze series like \(\sum_{n=1}^{\infty} \frac{n^2 - 1}{2n^2 + 3}\) and \(\sum_{n=2}^{\infty} \frac{3^n n^2}{n!}\) step by step.

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