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To solve this exercise, we need to prove the inequality by **mathematical induction**:

\[
\sum_{k=1}^{n} \frac{1}{k} \leq \frac{n+1}{2}
\]
for all \( n \in \mathbb{N}^* \) (the set of positive integers).

### Step 1: **Base Case** \( n = 1 \)

For \( n = 1 \), the left-hand side is:
\[
\sum_{k=1}^{1} \frac{1}{k} = \frac{1}{1} = 1
\]
The right-hand side is:
\[
\frac{1+1}{2} = \frac{2}{2} = 1
\]
Thus, for \( n = 1 \), we have:
\[
1 \leq 1
\]
which is true. Therefore, the base case holds.

### Step 2: **Inductive Hypothesis**

Assume that the inequality holds for some \( n = p \), i.e., we assume:
\[
\sum_{k=1}^{p} \frac{1}{k} \leq \frac{p+1}{2}
\]
This is the **inductive hypothesis**.

### Step 3: **Inductive Step**

We must prove that the inequality holds for \( n = p+1 \), i.e., we need to show:
\[
\sum_{k=1}^{p+1} \frac{1}{k} \leq \frac{p+2}{2}
\]

From the inductive hypothesis, we have:
\[
\sum_{k=1}^{p+1} \frac{1}{k} = \sum_{k=1}^{p} \frac{1}{k} + \frac{1}{p+1}
\]
Using the inductive hypothesis \( \sum_{k=1}^{p} \frac{1}{k} \leq \frac{p+1}{2} \), we can write:
\[
\sum_{k=1}^{p+1} \frac{1}{k} \leq \frac{p+1}{2} + \frac{1}{p+1}
\]
Now, we need to show that:
\[
\frac{p+1}{2} + \frac{1}{p+1} \leq \frac{p+2}{2}
\]
To do this, subtract \( \frac{p+1}{2} \) from both sides:
\[
\frac{1}{p+1} \leq \frac{1}{2}
\]
This inequality is true because \( \frac{1}{p+1} \leq \frac{1}{2} \) for all \( p \geq 1 \).

### Conclusion:

By the principle of mathematical induction, the inequality holds for all \( n \in \mathbb{N}^* \). Hence, we have proven that:
\[
\sum_{k=1}^{n} \frac{1}{k} \leq \frac{n+1}{2}
\]
for all \( n \in \mathbb{N}^* \).

Would you like more details on any step or have further questions?

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Here are 5 relative questions to expand your understanding:
1. How does mathematical induction work in proving inequalities?
2. Why is it necessary to check the base case in induction proofs?
3. What role does the inductive hypothesis play in an induction proof?
4. Can you think of other sums or inequalities where induction might apply?
5. How can we extend this method to more complex inequalities?

**Tip:** When proving inequalities using induction, always check the critical point where the inequality might hold equality—this helps ensure you understand the transition between steps.

Prove by induction that: for all n in natural numbers, the sum from k=1 to n of 1/k is less than or equal to (n+1)/2.

This page provides a step-by-step solution for proving the inequality ∑(1/k) ≤ (n+1)/2 using mathematical induction. Suitable for students in Grades 10-12, this problem involves applying the principle of induction to prove an inequality involving a sum of fractions.

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