Cho A = 1/101 + 1/102 + 1/103 + ... + 1/199 + 1/200. Chứng minh (1/2 < A < 1).

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Cho \[A = \frac{1}{{101}} + \frac{1}{{102}} + \frac{1}{{103}} + ... + \frac{1}{{199}} + \frac{1}{{200}}\]. Chứng minh \(\frac{1}{2} < A < 1.\)

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Giải thích

Hướng dẫn giải:

Ta có:

\(\frac{1}{{200}} < \frac{1}{{101}} < \frac{1}{{100}}\)

\(\frac{1}{{200}} < \frac{1}{{102}} < \frac{1}{{100}}\)

\(\frac{1}{{200}} < \frac{1}{{103}} < \frac{1}{{100}}\)

…

\(\frac{1}{{200}} < \frac{1}{{199}} < \frac{1}{{100}}\)

Suy ra:

\(\frac{1}{{200}} + \frac{1}{{200}} + \frac{1}{{200}} + ... + \frac{1}{{200}} < \frac{1}{{101}} + \frac{1}{{102}} + \frac{1}{{103}} + ... + \frac{1}{{199}} < \frac{1}{{100}} + \frac{1}{{100}} + \frac{1}{{100}} + ... + \frac{1}{{100}}\)

Hay \[99.\frac{1}{{200}} < \frac{1}{{101}} + \frac{1}{{102}} + \frac{1}{{103}} + ... + \frac{1}{{199}} < 99.\frac{1}{{100}}\]

\(\frac{{99}}{{200}} + \frac{1}{{200}} < \frac{1}{{101}} + \frac{1}{{102}} + \frac{1}{{103}} + ... + \frac{1}{{199}} + \frac{1}{{200}} < \frac{{99}}{{100}} + \frac{1}{{100}}\)

\(\frac{{100}}{{200}} < \frac{1}{{101}} + \frac{1}{{102}} + \frac{1}{{103}} + ... + \frac{1}{{199}} + \frac{1}{{200}} < \frac{{100}}{{100}}\)

Do đó \(\frac{{100}}{{200}} < A < \frac{{100}}{{100}}\)

Suy ra \(\frac{1}{2} < A < 1.\)