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The image you uploaded contains a series of problems involving sequences and series, each requiring to determine the general term formula and analyze convergence or divergence. Let's address each item sequentially:

### 1. Sequence: \( 1, \frac{2}{2}, \frac{3}{4}, \frac{4}{8}, \frac{5}{16}, \ldots \)
This sequence appears to have a general term \( a_n = \frac{n}{2^{n-1}} \).

#### Convergence/Divergence:
To determine if the sequence converges or diverges, consider the limit as \( n \) approaches infinity:
\[
\lim_{n \to \infty} \frac{n}{2^{n-1}} = 0
\]
since the exponential growth in the denominator outpaces the linear growth in the numerator. Thus, the sequence converges to 0.

### 2. Series: \( \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \ldots \)
The general term is \( \frac{n}{2^n} \).

#### Convergence/Divergence:
We use the ratio test to determine convergence:
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} \right| = \frac{1}{2}
\]
Since the limit is less than 1, the series converges by the ratio test.

### 3. Sequence: \( \frac{2}{1}, \frac{6}{3}, \frac{24}{5}, \ldots \)
This sequence seems to follow the pattern \( a_n = \frac{2^n \cdot n}{2n-1} \).

#### Convergence/Divergence:
The sequence diverges as \( n \to \infty \) because the numerator grows much faster than the denominator.

### 4. Series:
#### a. \(\sum_{n=1}^\infty (-1)^{n-1} \frac{n^2}{2^n}\)
#### b. \(\sum_{n=1}^\infty (-1)^{n+1} \frac{3^n}{n!}\)
#### c. \(\sum_{n=1}^\infty (-1)^{n+1} \frac{n^2}{n^3+4}\)
#### d. \(\sum_{n=2}^\infty (-1)^n \frac{1}{\sqrt{n}}\)

#### Convergence Analysis:
- a and b are both alternating series. For a, the terms \( \frac{n^2}{2^n} \) decrease in size and approach zero, hence it converges by the Alternating Series Test. For b, the terms \( \frac{3^n}{n!} \) also decrease rapidly (factorial growth in the denominator), so it converges.
- c is also an alternating series with terms \( \frac{n^2}{n^3+4} \) decreasing and approaching zero, so it converges.
- d, however, is an alternating series where the terms \( \frac{1}{\sqrt{n}} \) decrease but do not satisfy the condition for the Alternating Series Test (the decrease is not fast enough), so it diverges.

If you need detailed solutions or have further questions, let me know! Here are a few related questions you might find interesting:
1. How does the convergence of \( \sum_{n=1}^\infty \frac{n}{2^n} \) compare to that of \( \sum_{n=1}^\infty \frac{n^2}{2^n} \)?
2. Can you explain why the factorial in the denominator causes \( \sum_{n=1}^\infty \frac{3^n}{n!} \) to converge so rapidly?
3. What is the impact of the +4 in the denominator on the convergence of \( \sum_{n=1}^\infty \frac{n^2}{n^3+4} \)?
4. What conditions must a sequence satisfy for an alternating series to converge?
5. How does the ratio test confirm the convergence of the series \( \sum_{n=1}^\infty \frac{n}{2^n} \)?

**Tip:** When checking the convergence of a series, always consider the behavior of its terms as \( n \) approaches infinity; this often gives a clear indication of the series' behavior.

Several problems involving sequences and series, focusing on determining the general term and analyzing convergence or divergence.

This article walks through problems involving sequences and series, with a focus on determining the general term formulas and analyzing convergence or divergence. It covers the ratio test and alternating series test in detail. Suitable for students in advanced high school or university calculus.

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