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.