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.