Chứng minh rằng 1/2^2 +1/3^2 + 1/4^2 + ... + 1/2022^2 < 1

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Chứng minh \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < 1\).

Nhập câu trả lời ở đây

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Giải thích

Ta có: \(\frac{1}{{{2^2}}} < \frac{1}{{1.2}}\); \(\frac{1}{{{3^2}}} < \frac{1}{{2.3}}\); \(\frac{1}{{{4^2}}} < \frac{1}{{3.4}}\);…; \(\frac{1}{{{{2021}^2}}} < \frac{1}{{2020.2021}}\).

Đặt \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}}\)

\( = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2020.2021}}\)

\( \Rightarrow A < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2020}} - \frac{1}{{2021}}\)

\( \Rightarrow A < 1 - \frac{1}{{2021}}\)

\( \Rightarrow A < \frac{{2020}}{{2021}}\).

Vì \[2020 < 2021\] nên \(\frac{{2020}}{{2021}} < 1\).

Do đó \(A < \frac{{2020}}{{2021}} < 1\).

Vậy \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < 1\) (đpcm).