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We are asked to prove by mathematical induction that:

\[
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} \leq 2 - \frac{1}{n}, \quad \text{for all } n \geq 1.
\]

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

For \(n = 1\), the left-hand side is:

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

The right-hand side is:

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

Thus, for \(n = 1\), both sides are equal, i.e., \(1 \leq 1\), so the base case holds.

### Step 2: Inductive Hypothesis

Assume the statement is true for some \(n = k\), i.e.,

\[
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{k^2} \leq 2 - \frac{1}{k}.
\]

### Step 3: Inductive Step

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

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

Using the inductive hypothesis, we know:

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

Adding \(\frac{1}{(k+1)^2}\) to both sides:

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

We need to show that:

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

Simplifying the inequality:

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

Rearranging terms:

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

We know that:

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

We now check if:

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

Cross-multiply:

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

Expanding both sides:

\[
k^2 + 2k + 1 \geq k^2 + k.
\]

This simplifies to:

\[
2k + 1 \geq k,
\]

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

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

### Conclusion

By the principle of mathematical induction, the given inequality is true for all \(n \geq 1\).

---

Would you like further details or have any questions about this proof? Here are five related questions you might find interesting:

1. How does the harmonic series differ from this summation?
2. Can you explain why we added \(\frac{1}{(k+1)^2}\) to both sides in the inductive step?
3. What is the importance of checking the base case in induction?
4. Can this method be applied to other forms of summations involving squares?
5. How does the final inequality simplify in the inductive step?

**Tip:** When proving inequalities using induction, always check that the final step maintains a logical consistency between the inductive hypothesis and the statement you're proving for \(n + 1\).

Prove by mathematical induction that for all n >= 1, the sum of the series 1/1^2 + 1/2^2 + 1/3^2 + ... + 1/n^2 is less than or equal to 2 - 1/n.

Learn how to use mathematical induction to prove that the sum of the series 1/1^2 + 1/2^2 + ... + 1/n^2 is less than or equal to 2 - 1/n for all n >= 1. This proof involves key concepts in mathematical induction and inequality handling.

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