Let's address the questions one by one:

Q1: The Divergence Theorem

You are asked to identify which statement is true. Let's analyze the options:

• Choice 1 (A): $a_n = \frac{1}{n^2}$

  The sequence $a_n = \frac{1}{n^2}$ converges to 0 as $n \to \infty$. Additionally, the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges because this is a p-series with $p = 2$, and $p > 1$ implies convergence.

• Choice 2 (B): $a_n = \frac{1}{n}$

  The sequence $a_n = \frac{1}{n}$ converges to 0, but the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, which diverges.

• Choice 3 (C): $a_n = (-1)^n$

  The sequence $a_n = (-1)^n$ does not converge because it oscillates between 1 and -1. However, the series $\sum_{n=1}^{\infty} (-1)^n$ is an alternating series and does converge (by the alternating series test).

• Choice 4 (D): $a_n = \frac{n+1}{n}$

  The sequence $a_n = \frac{n+1}{n} = 1 + \frac{1}{n}$ converges to 1. However, the series $\sum_{n=1}^{\infty} \frac{n+1}{n}$ diverges, since the terms don't approach 0 quickly enough.

• Choice 5 (E): All statements are false.

Answer: Choice 1 (A) is true, because the sequence $\frac{1}{n^2}$ converges, and the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ also converges.

Q2: Fourth partial sum of $a_n = \frac{n+1}{n}$

The formula for $a_n = \frac{n+1}{n} = 1 + \frac{1}{n}$.

The fourth partial sum is the sum of the first four terms of the sequence:

$S_4 = \sum_{n=1}^{4} \left(1 + \frac{1}{n}\right)$

Let's calculate:

$S_4 = \left(1 + \frac{1}{1}\right) + \left(1 + \frac{1}{2}\right) + \left(1 + \frac{1}{3}\right) + \left(1 + \frac{1}{4}\right)$ $S_4 = (2) + \left(1 + \frac{1}{2}\right) + \left(1 + \frac{1}{3}\right) + \left(1 + \frac{1}{4}\right)$ $S_4 = 2 + \frac{3}{2} + \frac{4}{3} + \frac{5}{4}$

Now sum these fractions:

$S_4 = 2 + 1.5 + 1.\overline{33} + 1.25 = 6.08\ldots$

The exact form as a fraction would require finding a common denominator, but for simplicity:

$S_4 = \frac{205}{36}$

Q3: Which series converge?

Let's analyze each series:

• Choice 1 (A): $\sum_{n=1}^{\infty} \frac{2n-1}{4n}$

  This series diverges because the terms do not approach 0 as $n \to \infty$.

• Choice 2 (B): $\sum_{n=1}^{\infty} \frac{1}{n^4}$

  This is a p-series with $p = 4$. Since $p > 1$, the series converges.

• Choice 3 (C): $\sum_{n=1}^{\infty} \frac{1}{(-4)^n}$

  This is a geometric series with common ratio $r = -\frac{1}{4}$. Since $|r| < 1$, the series converges.

• Choice 4 (D): $\sum_{n=1}^{\infty} \frac{4n-1}{2n}$

  The terms of this series approach 2 as $n \to \infty$, so the series diverges.

• Choice 5 (E): None of the above.

Answer: B and C converge.

Would you like more details on any specific part or have any other questions? Here are five related questions you can think about:

1. What is the Alternating Series Test, and when can it be applied?
2. How do you determine whether a p-series converges or diverges?
3. How do partial sums relate to the convergence of a series?
4. Can a series converge if the corresponding sequence diverges?
5. What are the conditions for a geometric series to converge?

Tip: Always check whether the terms of a series approach 0 when determining its convergence—it’s a necessary condition!