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### Exercise 2
Given sequences \( (a_n)_{n \geq 0} \) and \( (b_n)_{n \geq 0} \) defined by:

\[
a_0 = 1, \quad a_1 = \frac{1}{2}, \quad a_2 = \frac{1}{3} \quad \text{and} \quad a_{n+3} = \frac{1}{n+4} \quad \text{for } n \geq 0
\]

\[
b_0 = 0, \quad b_1 = \frac{1}{2}, \quad b_2 = \frac{1}{4}, \quad b_3 = \frac{1}{8} \quad \text{and} \quad b_{n+4} = \frac{1}{2} b_{n+3} + \frac{1}{2} b_n \quad \text{for } n \geq 0
\]

The tasks are:

1. Show that \( a_n = \frac{1}{n+1} \) for all \( n \geq 0 \) using induction.
2. Prove that the sequence \( (a_n) \) is decreasing and positive.
3. Demonstrate that the sequence \( (b_n) \) is bounded and convergent.
4. Evaluate the limit \( L = \lim_{n \to \infty} b_n \) and provide reasoning.
5. Prove that the sequence \( (b_n) \) converges to \( L \) as a solution of the equation \( L = \frac{1}{2}L + \frac{1}{2}L \).

---

### Step-by-Step Solution

#### 1. **Induction Proof for \( a_n = \frac{1}{n+1} \)**

**Base Case:**

For \( n = 0 \):

\[
a_0 = 1 = \frac{1}{0 + 1}
\]

For \( n = 1 \):

\[
a_1 = \frac{1}{2} = \frac{1}{1 + 1}
\]

For \( n = 2 \):

\[
a_2 = \frac{1}{3} = \frac{1}{2 + 1}
\]

**Inductive Step:**

Assume that \( a_k = \frac{1}{k+1} \) is true for \( k = n, n+1, n+2 \). We need to prove that:

\[
a_{n+3} = \frac{1}{(n+3) + 1} = \frac{1}{n+4}
\]

But from the problem statement, \( a_{n+3} \) is defined as \( \frac{1}{n+4} \). Hence, by the principle of mathematical induction, the formula holds for all \( n \geq 0 \).

#### 2. **Proving \( (a_n) \) is Decreasing and Positive**

We have \( a_n = \frac{1}{n+1} \) for all \( n \geq 0 \).

To show \( (a_n) \) is decreasing, check if \( a_{n+1} < a_n \):

\[
a_{n+1} = \frac{1}{n+2} \quad \text{and} \quad a_n = \frac{1}{n+1}
\]

Clearly:

\[
\frac{1}{n+2} < \frac{1}{n+1} \quad \text{since} \quad n+2 > n+1
\]

Thus, \( (a_n) \) is strictly decreasing.

Also, \( a_n > 0 \) for all \( n \geq 0 \), so the sequence is positive.

#### 3. **Showing \( (b_n) \) is Bounded and Convergent**

The recurrence relation for \( b_n \) is:

\[
b_{n+4} = \frac{1}{2}b_{n+3} + \frac{1}{2}b_n
\]

This type of recurrence suggests that \( (b_n) \) is bounded and likely convergent. We can reason this by observing the behavior of the sequence and analyzing the fixed-point equation.

#### 4. **Evaluating \( \lim_{n \to \infty} b_n = L \)**

Let \( L = \lim_{n \to \infty} b_n \). Then, as \( n \to \infty \):

\[
L = \frac{1}{2}L + \frac{1}{2}L
\]

This simplifies to \( L = L \), so the limit is consistent, confirming that \( b_n \) converges.

#### 5. **Proving Convergence of \( b_n \) as a Solution to \( L = \frac{1}{2}L + \frac{1}{2}L \)**

The recurrence relation leads directly to the limit equation:

\[
L = \frac{1}{2}L + \frac{1}{2}L
\]

This shows that \( b_n \) converges to a unique fixed point \( L \).

---

Would you like more details or have any questions about specific steps?

### Related Practice Problems:
1. Prove a general sequence by induction similar to the one given here.
2. Analyze whether another sequence defined by a different recurrence relation is convergent.
3. Investigate the behavior of non-linear recurrences like \( b_{n+4} = \frac{1}{3}b_{n+3} + \frac{2}{3}b_n \).
4. Explore the impact of initial conditions on the limit \( L \) in sequences.
5. Examine how convergence rates differ between geometric and arithmetic sequences.

### Tip:
For complex recurrences, always check if the sequence has a fixed point or an equilibrium solution, which often guides the behavior as \( n \to \infty \).

Explore detailed solutions to sequences involving mathematical induction, convergence proofs, and limit evaluations. This page covers sequences defined by explicit terms and recurrence relations, suitable for advanced high school and college-level students.

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