Ex. 4 Using induction, show that
∑n 1 ≤ 2 − 1 i=1 i2 n

for all n ∈ N \ {0}.

We want to prove by induction that:

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

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

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

First, we check the base case when \( n = 1 \). The left-hand side of the inequality becomes:

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

Now check the right-hand side for \( n = 1 \):

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

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

\[
1 \leq 1,
\]

which holds true. Therefore, the base case is true.

### Step 2: Induction Hypothesis

Assume the inequality holds for some \( n = k \), i.e.,

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

### Step 3: Inductive Step

We now need to prove that the inequality also holds for \( n = k + 1 \), i.e., 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 for \( n = k + 1 \):

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

Using the induction hypothesis, we know:

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

Thus, we can write:

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

We need to show that:

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

This simplifies to proving:

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

Rearranging this inequality:

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

Simplifying the left-hand side:

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

Thus, the inequality becomes:

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

Multiplying both sides by \( (k+1)^2 \) to clear the denominator:

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

which simplifies to:

\[
\frac{k+1}{k} \geq 1,
\]

which is clearly true for all \( k \geq 1 \).

### Conclusion:

Since both the base case and the inductive step have been verified, by mathematical induction, the inequality

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

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

Would you like any further details or clarifications?

Here are 5 related questions:
1. How does the sum of squares compare to other series, such as harmonic or arithmetic series?
2. Can we apply a similar inductive proof for sums involving higher powers, like \( \sum \frac{1}{i^3} \)?
3. What are the implications of this inequality in terms of approximation of the sum?
4. How would this proof change if the domain were extended to include \( n = 0 \)?
5. How does this inequality relate to the convergence of infinite series?

**Tip**: Induction is especially useful for proving statements involving sums, sequences, and inequalities over the natural numbers.

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

This guide provides a step-by-step induction proof to show that the sum of reciprocal squares (∑ from i=1 to n of 1/i^2) is less than or equal to 2 - 1/n for all n in N \ {0}. Explore key concepts such as mathematical induction, series, and inequalities. Suitable for undergraduate students.

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