Cho \(S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}} + \frac{1}{{17}} + \frac{1}{{18}} + \frac{1}{{19}} + \frac{1}{{20}}\)
Hãy so sánh S và \(\frac{1}{2}\)
Ta có:
\(\begin{array}{l}
\frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}} + \frac{1}{{20}}\\
= \frac{{10}}{{20}} = \frac{1}{2}
\end{array}\)
Mà \(\frac{1}{{11}} > \frac{1}{{20}};\frac{1}{{12}} > \frac{1}{{20}}...;\frac{1}{{18}} > \frac{1}{{20}};\frac{1}{{19}} > \frac{1}{{20}}\)
Nên \(S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}} + \frac{1}{{17}} + \frac{1}{{18}} + \frac{1}{{19}} + \frac{1}{{20}}\) > \(\frac{1}{2}\)
-- Mod Toán 6
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