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.