We are given two sequences $(U_n)_{n \in \mathbb{N}}$ and $(V_n)_{n \in \mathbb{N}}$ defined by the relations: $U_0 = -\frac{3}{2}$ and $U_{n+1} = \frac{2}{3}U_n - 1$ $V_n = 2U_n + a$ 1. Determine the value of $a$ for which the sequence $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence. Specify its common ratio $q$ and its first term $V_0$.

1. Problem Description

We are given two sequences $(U_n)_{n \in \mathbb{N}}$ and $(V_n)_{n \in \mathbb{N}}$ defined by the relations:

$U_0 = -\frac{3}{2}$ and $U_{n+1} = \frac{2}{3}U_n - 1$

$V_n = 2U_n + a$

1. Determine the value of $a$ for which the sequence $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence. Specify its common ratio $q$ and its first term $V_0$.

2. Set $a = 6$. Express $V_n$ and $U_n$ as functions of $n$.

3. Show that the sequence $(U_n)_{n \in \mathbb{N}}$ is decreasing on $\mathbb{N}$.

4. Express the following sums as functions of $n$: $S_n = V_0 + V_1 + V_2 + \dots + V_n$ and $S'_n = U_0 + U_1 + U_2 + \dots + U_n$.

5. Study the convergence of the sequence $(S'_n)_{n \in \mathbb{N}}$.

6. Solution Steps

7. For $(V_n)$ to be geometric, we need $V_{n+1} = qV_n$ for some constant $q$.

$V_{n+1} = 2U_{n+1} + a = 2(\frac{2}{3}U_n - 1) + a = \frac{4}{3}U_n - 2 + a$

$qV_n = q(2U_n + a) = 2qU_n + aq$

Equating the two expressions for $V_{n+1} = qV_n$:

$\frac{4}{3}U_n - 2 + a = 2qU_n + aq$

For this equation to hold for all $n$, the coefficients of $U_n$ must be equal, and the constant terms must be equal.

$\frac{4}{3} = 2q$ which implies $q = \frac{2}{3}$

$-2 + a = aq = a(\frac{2}{3})$ which implies $-2 + a = \frac{2}{3}a$

$\frac{1}{3}a = 2$ which implies $a = 6$

$V_0 = 2U_0 + a = 2(-\frac{3}{2}) + 6 = -3 + 6 = 3$

2. With $a = 6$, we have $V_n = 2U_n + 6$ and $U_{n+1} = \frac{2}{3}U_n - 1$. Since we have shown that $V_n$ is a geometric sequence, we have $V_n = V_0 q^n = 3(\frac{2}{3})^n$.

$V_n = 2U_n + 6 = 3(\frac{2}{3})^n$

$2U_n = 3(\frac{2}{3})^n - 6$

$U_n = \frac{3}{2}(\frac{2}{3})^n - 3$

3. To show that $(U_n)$ is decreasing, we must show that $U_{n+1} - U_n < 0$ for all $n$.

$U_{n+1} - U_n = (\frac{3}{2}(\frac{2}{3})^{n+1} - 3) - (\frac{3}{2}(\frac{2}{3})^n - 3) = \frac{3}{2}(\frac{2}{3})^{n+1} - \frac{3}{2}(\frac{2}{3})^n = \frac{3}{2}(\frac{2}{3})^n(\frac{2}{3} - 1) = \frac{3}{2}(\frac{2}{3})^n(-\frac{1}{3}) = -\frac{1}{2}(\frac{2}{3})^n$

Since $(\frac{2}{3})^n > 0$ for all $n$, we have $U_{n+1} - U_n = -\frac{1}{2}(\frac{2}{3})^n < 0$. Therefore, $(U_n)$ is a decreasing sequence.

4. $S_n = V_0 + V_1 + \dots + V_n$. Since $V_n$ is a geometric sequence, we can use the formula for the sum of a geometric series:

$S_n = V_0 \frac{1 - q^{n+1}}{1 - q} = 3 \frac{1 - (\frac{2}{3})^{n+1}}{1 - \frac{2}{3}} = 3 \frac{1 - (\frac{2}{3})^{n+1}}{\frac{1}{3}} = 9(1 - (\frac{2}{3})^{n+1}) = 9 - 9(\frac{2}{3})^{n+1}$

$S'_n = U_0 + U_1 + \dots + U_n = \sum_{k=0}^n U_k = \sum_{k=0}^n (\frac{3}{2}(\frac{2}{3})^k - 3) = \frac{3}{2}\sum_{k=0}^n (\frac{2}{3})^k - \sum_{k=0}^n 3 = \frac{3}{2}\frac{1 - (\frac{2}{3})^{n+1}}{1 - \frac{2}{3}} - 3(n+1) = \frac{3}{2} \frac{1 - (\frac{2}{3})^{n+1}}{\frac{1}{3}} - 3n - 3 = \frac{9}{2}(1 - (\frac{2}{3})^{n+1}) - 3n - 3 = \frac{9}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n - 3 = \frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n$

5. To study the convergence of the sequence $(S'_n)_{n \in \mathbb{N}}$, we look at the limit as $n$ approaches infinity:

$\lim_{n \to \infty} S'_n = \lim_{n \to \infty} (\frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n)$

Since $\lim_{n \to \infty} (\frac{2}{3})^{n+1} = 0$, the term $\frac{9}{2}(\frac{2}{3})^{n+1}$ goes to 0 as $n$ approaches infinity. However, the term $-3n$ goes to $-\infty$ as $n$ approaches infinity. Therefore, $\lim_{n \to \infty} S'_n = -\infty$.

The sequence $(S'_n)_{n \in \mathbb{N}}$ diverges to $-\infty$.

6. Final Answer

7. $a = 6$, $q = \frac{2}{3}$, $V_0 = 3$.

8. $V_n = 3(\frac{2}{3})^n$, $U_n = \frac{3}{2}(\frac{2}{3})^n - 3$.

9. The sequence $(U_n)$ is decreasing.

1

0. $S_n = 9 - 9(\frac{2}{3})^{n+1}$, $S'_n = \frac{3}{2} - \frac{9}{2}(\frac{2}{3})^{n+1} - 3n$.

1

1. The sequence $(S'_n)$ diverges to $-\infty$.