The problem defines a sequence $f$ such that for all $n \in \mathbb{N}$, $2f(n+1) + f(n) = 0$, and $f(0) \neq 0$. 1. Prove that for any integer $k \ge 2$, $f(0)f(k) = f(2)f(k-2)$.

1. Problem Description

The problem defines a sequence $f$ such that for all $n \in \mathbb{N}$, $2f(n+1) + f(n) = 0$, and $f(0) \neq 0$.

1. Prove that for any integer $k \ge 2$, $f(0)f(k) = f(2)f(k-2)$.

2. a) Calculate $f(n)$ as a function of $f(0)$. Deduce $f(n)$ as a function of $n$ knowing that $f(0) = 1$.

b) Calculate in this case the sum $S_n = f(0) + f(1) + f(2) + \dots + f(n)$.

3. Study the convergence of $(S_n)$.

2. Solution Steps

1. We want to prove that for any integer $k \ge 2$, $f(0)f(k) = f(2)f(k-2)$.

From the definition of the sequence, we have $2f(n+1) + f(n) = 0$, which implies $f(n+1) = -\frac{1}{2} f(n)$.

This means that $f$ is a geometric sequence with ratio $-\frac{1}{2}$. Therefore, $f(n) = f(0) \left(-\frac{1}{2}\right)^n$ for all $n \in \mathbb{N}$.

Now, let's consider $f(0)f(k)$ and $f(2)f(k-2)$:

$f(0)f(k) = f(0) \cdot f(0) \left(-\frac{1}{2}\right)^k = f(0)^2 \left(-\frac{1}{2}\right)^k$.

$f(2)f(k-2) = f(0) \left(-\frac{1}{2}\right)^2 \cdot f(0) \left(-\frac{1}{2}\right)^{k-2} = f(0)^2 \left(-\frac{1}{2}\right)^2 \left(-\frac{1}{2}\right)^{k-2} = f(0)^2 \left(-\frac{1}{2}\right)^{2+k-2} = f(0)^2 \left(-\frac{1}{2}\right)^k$.

Since $f(0)f(k) = f(0)^2 \left(-\frac{1}{2}\right)^k = f(2)f(k-2)$, the statement is proven.

2. a) Calculate $f(n)$ as a function of $f(0)$.

As we have seen earlier, $f(n) = f(0) \left(-\frac{1}{2}\right)^n$.

Now, we deduce $f(n)$ as a function of $n$ knowing that $f(0) = 1$.

So, $f(n) = 1 \cdot \left(-\frac{1}{2}\right)^n = \left(-\frac{1}{2}\right)^n$.

b) Calculate in this case the sum $S_n = f(0) + f(1) + f(2) + \dots + f(n)$.

$S_n = \sum_{i=0}^{n} f(i) = \sum_{i=0}^{n} \left(-\frac{1}{2}\right)^i$.

This is a geometric series with first term $a = 1$ and ratio $r = -\frac{1}{2}$.

The sum of the first $n+1$ terms is given by the formula:

$S_n = \frac{a(1-r^{n+1})}{1-r} = \frac{1(1 - (-\frac{1}{2})^{n+1})}{1 - (-\frac{1}{2})} = \frac{1 - (-\frac{1}{2})^{n+1}}{1 + \frac{1}{2}} = \frac{1 - (-\frac{1}{2})^{n+1}}{\frac{3}{2}} = \frac{2}{3}\left(1 - \left(-\frac{1}{2}\right)^{n+1}\right)$.

3. Study the convergence of $(S_n)$.

We need to find the limit of $S_n$ as $n$ approaches infinity:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{2}{3}\left(1 - \left(-\frac{1}{2}\right)^{n+1}\right)$.

Since $\lim_{n \to \infty} \left(-\frac{1}{2}\right)^{n+1} = 0$, we have:

$\lim_{n \to \infty} S_n = \frac{2}{3}(1 - 0) = \frac{2}{3}$.

Therefore, the sequence $(S_n)$ converges to $\frac{2}{3}$.

3. Final Answer

1. Proven that for any integer $k \ge 2$, $f(0)f(k) = f(2)f(k-2)$.

2. a) $f(n) = f(0)(-\frac{1}{2})^n$. If $f(0) = 1$, then $f(n) = (-\frac{1}{2})^n$.

b) $S_n = \frac{2}{3}(1 - (-\frac{1}{2})^{n+1})$.

3. The sequence $(S_n)$ converges to $\frac{2}{3}$.