4 Using induction, show that
n
∑
i=1
1
i
2 ≤ 2 −
1
n
for all n ∈ N \ {0}.

We are tasked with proving the inequality:

\[
\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}
\]

for all \(n \in \mathbb{N} \setminus \{0\}\), using the principle of **mathematical induction**.

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

When \(n = 1\), the left-hand side becomes:

\[
\sum_{i=1}^{1} \frac{1}{i^2} = \frac{1}{1^2} = 1.
\]

The right-hand side becomes:

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

So for \(n = 1\), we have:

\[
1 \leq 1,
\]

which is true. Hence, the base case holds.

### Step 2: Inductive Hypothesis

Assume that for some \(n = k\), the inequality holds. That is, assume:

\[
\sum_{i=1}^{k} \frac{1}{i^2} \leq 2 - \frac{1}{k}.
\]

### Step 3: Inductive Step

We need to prove that the inequality holds for \(n = k + 1\). That is, we need to show:

\[
\sum_{i=1}^{k+1} \frac{1}{i^2} \leq 2 - \frac{1}{k+1}.
\]

Start with the left-hand side:

\[
\sum_{i=1}^{k+1} \frac{1}{i^2} = \sum_{i=1}^{k} \frac{1}{i^2} + \frac{1}{(k+1)^2}.
\]

By the inductive hypothesis, we know:

\[
\sum_{i=1}^{k} \frac{1}{i^2} \leq 2 - \frac{1}{k}.
\]

Thus:

\[
\sum_{i=1}^{k+1} \frac{1}{i^2} \leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2}.
\]

Now, we need to show that:

\[
2 - \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1}.
\]

Simplifying the right-hand side, we get:

\[
2 - \frac{1}{k+1}.
\]

Thus, we want to prove that:

\[
-\frac{1}{k} + \frac{1}{(k+1)^2} \leq -\frac{1}{k+1}.
\]

This simplifies to:

\[
\frac{1}{(k+1)^2} \leq \frac{1}{k} - \frac{1}{k+1}.
\]

Simplifying the right-hand side:

\[
\frac{1}{k} - \frac{1}{k+1} = \frac{k+1 - k}{k(k+1)} = \frac{1}{k(k+1)}.
\]

Thus, we need to show:

\[
\frac{1}{(k+1)^2} \leq \frac{1}{k(k+1)}.
\]

Cross-multiplying (which is valid since all terms are positive), we get:

\[
k(k+1) \leq (k+1)^2.
\]

Simplifying:

\[
k(k+1) \leq (k+1)(k+1),
\]

which simplifies to:

\[
k(k+1) \leq k^2 + 2k + 1,
\]

which is true because:

\[
k(k+1) = k^2 + k \quad \text{and} \quad k^2 + k \leq k^2 + 2k + 1.
\]

Therefore, the inequality holds, and the inductive step is complete.

### Conclusion

By the principle of mathematical induction, we have shown that:

\[
\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}
\]

for all \(n \in \mathbb{N} \setminus \{0\}\).

---

Would you like further clarifications or details on this proof?

Here are 5 questions related to the topic:
1. How does mathematical induction work in general?
2. Why is the base case important in mathematical induction?
3. Can you apply this induction method to prove other types of summation inequalities?
4. What are common mistakes to avoid when using induction?
5. How does this inequality relate to the harmonic series?

**Tip**: In induction proofs, always clearly define the base case and inductive hypothesis before moving to the inductive step.

Using induction, show that ∑(i=1 to n) 1/i^2 ≤ 2 − 1/n for all n ∈ N \ {0}.

Learn how to use mathematical induction to prove the inequality ∑(i=1 to n) 1/i^2 ≤ 2 − 1/n. This detailed step-by-step guide explains key concepts in series summation and inequalities, suitable for advanced high school or undergraduate students.

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