We are given a sequence $(U_n)_{n \in \mathbb{N}}$ defined by $U_0 = 1$ and $U_{n+1} = \frac{1}{2} U_n + \frac{1}{4}$. We also define a sequence $(V_n)_{n \in \mathbb{N}}$ by $V_n = U_n - \frac{1}{2}$. The problem asks us to: 1. Show that $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence and find its ratio $q$ and first term $V_0$.

1. Problem Description

We are given a sequence $(U_n)_{n \in \mathbb{N}}$ defined by $U_0 = 1$ and $U_{n+1} = \frac{1}{2} U_n + \frac{1}{4}$. We also define a sequence $(V_n)_{n \in \mathbb{N}}$ by $V_n = U_n - \frac{1}{2}$. The problem asks us to:

1. Show that $(V_n)_{n \in \mathbb{N}}$ is a geometric sequence and find its ratio $q$ and first term $V_0$.

2. Express $V_n$ and $U_n$ as functions of $n$.

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

4. Express $S_n = V_0 + V_1 + \dots + V_n$ and $S'_n = U_0 + U_1 + \dots + U_n$ as functions of $n$.

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

2. Solution Steps

1. Geometric Sequence:

We want to show that $V_{n+1} = q V_n$ for some constant $q$.

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

Therefore, $V_{n+1} = \frac{1}{2} V_n$. This shows that $(V_n)$ is a geometric sequence with ratio $q = \frac{1}{2}$.

The first term is $V_0 = U_0 - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}$.

2. Express $V_n$ and $U_n$ in terms of $n$:

Since $(V_n)$ is a geometric sequence with first term $V_0 = \frac{1}{2}$ and ratio $q = \frac{1}{2}$, we have:

$V_n = V_0 q^n = \frac{1}{2} (\frac{1}{2})^n = (\frac{1}{2})^{n+1}$.

Since $V_n = U_n - \frac{1}{2}$, we have $U_n = V_n + \frac{1}{2} = (\frac{1}{2})^{n+1} + \frac{1}{2}$.

3. Show $(U_n)$ is decreasing:

We need to show that $U_{n+1} < U_n$ for all $n \in \mathbb{N}$.

$U_{n+1} = (\frac{1}{2})^{n+2} + \frac{1}{2}$ and $U_n = (\frac{1}{2})^{n+1} + \frac{1}{2}$.

$U_n - U_{n+1} = (\frac{1}{2})^{n+1} + \frac{1}{2} - ((\frac{1}{2})^{n+2} + \frac{1}{2}) = (\frac{1}{2})^{n+1} - (\frac{1}{2})^{n+2} = (\frac{1}{2})^{n+1} (1 - \frac{1}{2}) = (\frac{1}{2})^{n+1} (\frac{1}{2}) = (\frac{1}{2})^{n+2} > 0$.

Since $U_n - U_{n+1} > 0$, we have $U_n > U_{n+1}$, which means $(U_n)$ is decreasing.

4. Express $S_n$ and $S'_n$ in terms of $n$:

$S_n = V_0 + V_1 + \dots + V_n = \sum_{k=0}^{n} V_k$. This is a geometric series with first term $V_0 = \frac{1}{2}$, ratio $q = \frac{1}{2}$, and $n+1$ terms.

The sum of a geometric series is given by:

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

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

$\sum_{k=0}^{n} (\frac{1}{2})^{k+1} = \frac{1}{2} + \frac{1}{4} + \dots + (\frac{1}{2})^{n+1} = S_n = 1 - (\frac{1}{2})^{n+1}$.

$\sum_{k=0}^{n} \frac{1}{2} = (n+1) \frac{1}{2} = \frac{n+1}{2}$.

So, $S'_n = 1 - (\frac{1}{2})^{n+1} + \frac{n+1}{2}$.

5. Convergence of $(S_n)$ and $(S'_n)$:

As $n \to \infty$, $(\frac{1}{2})^{n+1} \to 0$.

Therefore, $\lim_{n \to \infty} S_n = \lim_{n \to \infty} (1 - (\frac{1}{2})^{n+1}) = 1 - 0 = 1$. Thus, $(S_n)$ converges to

1. $\lim_{n \to \infty} S'_n = \lim_{n \to \infty} (1 - (\frac{1}{2})^{n+1} + \frac{n+1}{2}) = 1 - 0 + \lim_{n \to \infty} \frac{n+1}{2} = \infty$. Thus, $(S'_n)$ diverges to $\infty$.

3. Final Answer

1. The sequence $(V_n)$ is geometric with ratio $q = \frac{1}{2}$ and first term $V_0 = \frac{1}{2}$.

2. $V_n = (\frac{1}{2})^{n+1}$ and $U_n = (\frac{1}{2})^{n+1} + \frac{1}{2}$.

3. The sequence $(U_n)$ is decreasing on $\mathbb{N}$.

4. $S_n = 1 - (\frac{1}{2})^{n+1}$ and $S'_n = 1 - (\frac{1}{2})^{n+1} + \frac{n+1}{2}$.

5. The sequence $(S_n)$ converges to 1 and the sequence $(S'_n)$ diverges to $\infty$.