The problem requires us to find the first five terms of the given sequences $\{a_n\}$, determine if each sequence converges or diverges, and if it converges, find its limit as $n$ approaches infinity. We will solve problem number 2: $a_n = \frac{3n+2}{n+1}$.

1. Problem Description

The problem requires us to find the first five terms of the given sequences $\{a_n\}$, determine if each sequence converges or diverges, and if it converges, find its limit as $n$ approaches infinity. We will solve problem number 2: $a_n = \frac{3n+2}{n+1}$.

2. Solution Steps

First, we find the first five terms of the sequence by substituting $n=1, 2, 3, 4, 5$ into the formula for $a_n$:

$a_1 = \frac{3(1)+2}{1+1} = \frac{5}{2} = 2.5$

$a_2 = \frac{3(2)+2}{2+1} = \frac{8}{3} \approx 2.67$

$a_3 = \frac{3(3)+2}{3+1} = \frac{11}{4} = 2.75$

$a_4 = \frac{3(4)+2}{4+1} = \frac{14}{5} = 2.8$

$a_5 = \frac{3(5)+2}{5+1} = \frac{17}{6} \approx 2.83$

Next, we find the limit of the sequence as $n$ approaches infinity:

$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{3n+2}{n+1}$

To find this limit, we can divide both the numerator and the denominator by $n$:

$\lim_{n\to\infty} \frac{3n+2}{n+1} = \lim_{n\to\infty} \frac{\frac{3n}{n}+\frac{2}{n}}{\frac{n}{n}+\frac{1}{n}} = \lim_{n\to\infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}}$

As $n$ approaches infinity, $\frac{2}{n}$ and $\frac{1}{n}$ both approach

0. Therefore,

$\lim_{n\to\infty} \frac{3+\frac{2}{n}}{1+\frac{1}{n}} = \frac{3+0}{1+0} = \frac{3}{1} = 3$

Since the limit exists and is equal to 3, the sequence converges to

3.

3. Final Answer

The first five terms of the sequence are $2.5, \frac{8}{3}, \frac{11}{4}, \frac{14}{5}, \frac{17}{6}$.

The sequence converges, and $\lim_{n\to\infty} a_n = 3$.