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Test for convergence for question (d) and (e)

Let's tackle the convergence of the series from (d) and (e).

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**(d) \(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n + 3}\)**

This series is an **alternating series** because of the \((-1)^n\) factor, which makes the terms alternate between positive and negative. To test for convergence of an alternating series, we can use the **Alternating Series Test** (also called the **Leibniz criterion**), which states that the series \(\sum (-1)^n b_n\) converges if:

1. \(b_n\) is positive,
2. \(b_n\) is decreasing, and
3. \(\lim_{n \to \infty} b_n = 0\).

In this case, \(b_n = \frac{1}{4n + 3}\).

- \(b_n\) is positive for all \(n\).
- \(b_n = \frac{1}{4n + 3}\) is decreasing because the denominator increases as \(n\) increases.
- \(\lim_{n \to \infty} \frac{1}{4n + 3} = 0\).

Since all three conditions of the Alternating Series Test are satisfied, **the series converges**.

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**(e) \(\sum_{n=1}^{\infty} \frac{2n - 1}{\sqrt{n^3 + n + 2}}\)**

To determine the convergence of this series, we can use the **Limit Comparison Test**. This test is especially useful for series that are not easily recognized as geometric or p-series.

We first analyze the general behavior of the terms for large \(n\). For large \(n\), the term \(2n - 1 \approx 2n\) and \(\sqrt{n^3 + n + 2} \approx \sqrt{n^3} = n^{3/2}\). So, the general behavior of the terms is approximately:

\[
\frac{2n - 1}{\sqrt{n^3 + n + 2}} \sim \frac{2n}{n^{3/2}} = \frac{2}{n^{1/2}}.
\]

This suggests we compare the given series with the simpler series \(\sum \frac{1}{n^{1/2}}\). This is a **p-series** with \(p = \frac{1}{2}\), which is known to **diverge** because \(p \leq 1\).

Now, applying the **Limit Comparison Test**:

\[
\lim_{n \to \infty} \frac{\frac{2n - 1}{\sqrt{n^3 + n + 2}}}{\frac{1}{n^{1/2}}} = \lim_{n \to \infty} \frac{(2n - 1)n^{1/2}}{\sqrt{n^3 + n + 2}}.
\]

For large \(n\), this limit simplifies to:

\[
\lim_{n \to \infty} \frac{2n^{3/2}}{n^{3/2}} = 2.
\]

Since the limit is a positive finite number, by the Limit Comparison Test, the original series behaves like \(\sum \frac{1}{n^{1/2}}\), which diverges. Therefore, **the series diverges**.

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**Summary:**
- (d) \(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n + 3}\): **Converges** (by the Alternating Series Test).
- (e) \(\sum_{n=1}^{\infty} \frac{2n - 1}{\sqrt{n^3 + n + 2}}\): **Diverges** (by the Limit Comparison Test).

Let me know if you'd like further explanations or if you have any questions!

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Here are five related questions:
1. How does the p-series test determine convergence?
2. What is the difference between absolute and conditional convergence?
3. Can an alternating series ever diverge? If so, when?
4. How does the Limit Comparison Test differ from the Direct Comparison Test?
5. When would you use the Ratio Test to determine convergence?

**Tip**: Always simplify the terms of a series for large \(n\) to determine its behavior when applying convergence tests.

This page tests the convergence of two infinite series: an alternating series using the Alternating Series Test and a general series using the Limit Comparison Test. Explore key calculus concepts like alternating series, limit comparison, and p-series behavior. Suitable for advanced high school and introductory college students.

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