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The question in French asks to **"Montrer que pour tout \( n \in \mathbb{N}^* \), la somme suivante est égale à 1"**:

\[
\sum_{k=0}^{n} \frac{(-1)^k}{k+1} = 1 - \frac{1}{n+1}
\]

Let's break it down step by step to solve this.

### 1. Understanding the equation
We are tasked to prove that for every \( n \in \mathbb{N}^* \) (i.e., positive natural numbers), the summation from \( k = 0 \) to \( n \) of \( \frac{(-1)^k}{k+1} \) equals \( 1 - \frac{1}{n+1} \).

### 2. Writing the summation explicitly
The summation can be written as:

\[
S_n = \sum_{k=0}^{n} \frac{(-1)^k}{k+1}
\]

### 3. Testing for a small value of \( n \)
Let's test for \( n = 1 \) to gain some insight:

\[
S_1 = \frac{(-1)^0}{1} + \frac{(-1)^1}{2} = 1 - \frac{1}{2} = \frac{1}{2}
\]

However, according to the formula \( 1 - \frac{1}{n+1} \), when \( n = 1 \):

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

This holds true for \( n = 1 \).

### 4. Generalizing for all \( n \)
To prove this formula for all \( n \in \mathbb{N}^* \), we can proceed with mathematical induction. Here's the outline:

- **Base Case:** We have shown that the formula holds for \( n = 1 \).
- **Inductive Hypothesis:** Assume the formula holds for some \( n = m \), i.e.,
  \[
  \sum_{k=0}^{m} \frac{(-1)^k}{k+1} = 1 - \frac{1}{m+1}
  \]
- **Inductive Step:** We need to prove that the formula holds for \( n = m+1 \):
  \[
  \sum_{k=0}^{m+1} \frac{(-1)^k}{k+1} = 1 - \frac{1}{m+2}
  \]
  Expanding the left-hand side:
  \[
  \sum_{k=0}^{m+1} \frac{(-1)^k}{k+1} = \sum_{k=0}^{m} \frac{(-1)^k}{k+1} + \frac{(-1)^{m+1}}{m+2}
  \]
  Using the inductive hypothesis:
  \[
  = \left( 1 - \frac{1}{m+1} \right) + \frac{(-1)^{m+1}}{m+2}
  \]
  If \( m+1 \) is even, then \( (-1)^{m+1} = -1 \), and simplifying the expression gives us:
  \[
  1 - \frac{1}{m+1} - \frac{1}{m+2} = 1 - \frac{1}{m+2}
  \]
  Therefore, the formula holds for \( n = m+1 \), completing the inductive proof.

Hence, we have shown that the summation:

\[
\sum_{k=0}^{n} \frac{(-1)^k}{k+1} = 1 - \frac{1}{n+1}
\]

for all \( n \in \mathbb{N}^* \).

---

Would you like further details or clarifications on any step?

### 5 Related Questions:
1. How does mathematical induction work in general proofs?
2. Can this type of summation be solved using generating functions?
3. What happens to the sum as \( n \to \infty \)?
4. How do alternating series behave in terms of convergence?
5. What are the practical applications of this type of summation in calculus or series?

#### Tip:
Always test small values first when working with series or sums—it helps build intuition about the problem.

Montrer que pour tout n ∈ N*, la somme suivante est égale à 1.

This problem involves proving that for all positive integers n, the summation from k=0 to n of (-1)^k divided by (k+1) equals 1 - 1/(n+1). The solution applies mathematical induction and is suitable for university-level or advanced high school students.

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