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Ta có :
\(\frac{1}{13}< \frac{1}{12}\)
\(\frac{1}{14}< \frac{1}{12}\)
\(\frac{1}{15}< \frac{1}{12}\)
\(\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=3\cdot\frac{1}{12}=\frac{1}{4}\) (1)
Ta cũng có :
\(\frac{1}{61}< \frac{1}{60}\)
\(\frac{1}{62}< \frac{1}{60}\)
\(\frac{1}{63}< \frac{1}{60}\)
\(\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=3\cdot\frac{1}{60}=\frac{1}{20}\) (2)
Từ (1) ; (2) \(\Rightarrow S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
=> S < \(\frac{1}{2}\) (đpcm)
\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=0,36833......\)
mà \(\frac{1}{2}=0,5\)
\(0,36833..< 0,5\)
Vậy \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{2}\)
Đặt \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=A\)
Ta có : \(A=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{28}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{97}\right)\)
\(A< \frac{1}{5}\left(\frac{1}{14.3}\right)+\left(\frac{1}{61.3}\right)\)
\(A< \frac{1}{5}+\frac{3}{14}+\frac{3}{61}\)
\(A< \frac{1}{5}+\frac{3}{12}+\frac{1}{20}\)
\(A< \frac{1}{2}\left(ĐPM\right)\).