Ex. 4 Using induction, show that
∑n 1 ≤ 2 − 1 i=1 i2 n
∑n 1 √ √≥n
i
1
for all n ∈ N \ {0}.

To prove this inequality using **mathematical induction**, we need to show the following statement:

\[
\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}
\]
for all \( n \in \mathbb{N} \setminus \{0\} \).

### Step 1: Base Case
For \( n = 1 \), the left-hand side (LHS) is:

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

Now, calculate the right-hand side (RHS) for \( n = 1 \):

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

Since \( 1 = 1 \), the base case holds for \( n = 1 \).

### Step 2: Inductive Hypothesis
Assume the statement is true for some \( n = k \), i.e.,

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

We need to show that the statement also holds for \( n = k+1 \), i.e.,

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

### Step 3: Inductive Step
We begin with the inductive hypothesis:

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

Now, add \( \frac{1}{(k+1)^2} \) to both sides:

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

We need to prove that:

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

This simplifies to showing:

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

Rearranging this inequality:

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

We now simplify the left-hand side:

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

So, the inequality becomes:

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

Multiplying both sides by \( (k+1)^2 \) gives:

\[
k+1 \geq k
\]

This is true, since \( k+1 > k \) for all \( k \in \mathbb{N} \).

### 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 details on any part of the proof?

### Follow-up Questions:
1. What is the principle of mathematical induction, and how does it work?
2. How does the inequality change as \( n \) increases?
3. What is the intuition behind using induction for this kind of inequality?
4. Can you apply a similar method to other summations?
5. How do the sums involving \( 1/i^2 \) compare to sums involving \( 1/i \)?

### Tip:
When proving inequalities, try simplifying terms step by step to break down the problem logically.

Using induction, show that ∑n 1 ≤ 2 − 1 i=1 i2 n ∑n 1 √ √≥n i 1 for all n ∈ N \ {0}.

Explore how to prove the inequality ∑(1/i^2) ≤ 2 - 1/n for all n ∈ N\{0} using the principle of mathematical induction. This detailed solution walks through the base case, inductive hypothesis, and inductive step, applying summation and inequality concepts. Ideal for undergraduate-level mathematics students.

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