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The question is asking to show that the sequence \( v_n = \frac{5n}{n+1} \), defined for every natural number \( n \), is bounded above by 5.

To prove that \( v_n \) is bounded by 5, we need to show that for all \( n \in \mathbb{N} \), the inequality \( v_n \leq 5 \) holds true.

### Step-by-Step Proof:

1. **Sequence Definition:**
   \[
   v_n = \frac{5n}{n+1}
   \]

2. **Inequality to Prove:**
   We want to show:
   \[
   \frac{5n}{n+1} \leq 5
   \]

3. **Simplify the Inequality:**
   Multiply both sides of the inequality by \( n+1 \) (which is positive for all \( n \in \mathbb{N} \)) to avoid changing the inequality's direction:
   \[
   5n \leq 5(n + 1)
   \]
   This simplifies to:
   \[
   5n \leq 5n + 5
   \]

4. **Simplify Further:**
   Subtract \( 5n \) from both sides:
   \[
   0 \leq 5
   \]
   This is a true statement.

Since this inequality holds true for all \( n \in \mathbb{N} \), we conclude that:
\[
v_n \leq 5
\]
Thus, the sequence \( v_n = \frac{5n}{n+1} \) is indeed bounded above by 5.

Would you like any further details or have any other questions on this?

### Related Questions:
1. Can we determine the limit of the sequence \( v_n \) as \( n \to \infty \)?
2. How can we show if the sequence \( v_n \) is strictly increasing or decreasing?
3. What happens to \( v_n \) for small values of \( n \), such as \( n = 0, 1, 2 \)?
4. Is there a similar sequence that converges to a different value?
5. How do we graphically represent the behavior of \( v_n \) as \( n \) increases?

**Tip:** When dealing with inequalities involving sequences, multiplying by positive quantities allows you to keep the inequality direction intact.

Montrer que la suite (v_n) définie pour tout entier naturel n par v_n = 5n/(n+1) est majorée par 5.

Learn how to prove that the sequence v_n = 5n/(n+1), defined for all natural numbers n, is bounded above by 5. This problem demonstrates key algebraic techniques in working with sequences and inequalities, suitable for high school students.

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