Chứng minh S < 1.

Lời giải:

Ta có: \[S = \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} + ... + \frac{1}{{{{2025}^2}}}\]

\[S < \frac{1}{{{3^2} - 1}} + \frac{1}{{{5^2} - 1}} + \frac{1}{{{7^2} - 1}} + ... + \frac{1}{{{{2025}^2} - 1}}\]

\[S < \frac{1}{{\left( {3 - 1} \right)\left( {3 + 1} \right)}} + \frac{1}{{\left( {5 - 1} \right)\left( {5 + 1} \right)}} + ... + \frac{1}{{\left( {2025 - 1} \right)\left( {2025 + 1} \right)}}\]

\[S < \frac{1}{{2 \cdot 4}} + \frac{1}{{4 \cdot 6}} + ... + \frac{1}{{2024 \cdot 2026}}\]

\[S < \frac{1}{2} \cdot \left( {\frac{2}{{2 \cdot 4}} + \frac{2}{{4 \cdot 6}} + ... + \frac{2}{{2024 \cdot 2026}}} \right)\]

\[S < \frac{1}{2} \cdot \left( {\frac{1}{2} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + ... + \frac{1}{{2024}} - \frac{1}{{2026}}} \right)\]

\[S < \frac{1}{2} \cdot \left( {\frac{1}{2} - \frac{1}{{2026}}} \right) = \frac{1}{4} - \frac{1}{{4052}} < 1.\]

Vậy S < 1.