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Ta có: S = \(\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{10}{60}=\frac{1}{6}\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(\Rightarrow S>\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (1)
Lại có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\Rightarrow S< \frac{47}{60}< \frac{48}{60}=\frac{4}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}< S< \frac{4}{5}\) (đpcm)
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S có 30 số hạng. Nhóm thành 3 nhóm, mỗi nhóm 10 số hạng
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{42}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(S<\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(S<\frac{10}{30}+\frac{10}{40}+\frac{10}{50}\) ; \(S<\frac{47}{60}<\frac{48}{60}=\frac{4}{5}\) (1)
\(S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(S<\frac{10}{40}+\frac{10}{50}+\frac{10}{60}\) ; \(S>\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}\)<S<\(\frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
Ta có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{10}{60}\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (1)
Lại có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+...+\frac{1}{50}< \frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}< S< \frac{4}{5}\)
Đặt \(S = \frac{1}{31} + \frac{1}{32} + \frac{1}{33} + . . . + \frac{1}{59} + \frac{1}{60}\)
S có 30 số hạng.Nhóm thành ba nhóm, mỗi nhóm có 10 số hạng
\(S = \left(\right. \frac{1}{31} + \frac{1}{32} + . . . + \frac{1}{40} \left.\right) + \left(\right. \frac{1}{41} + \frac{1}{42} + \frac{1}{43} + . . . + \frac{1}{50} \left.\right) + \left(\right. \frac{1}{51} + \frac{1}{52} + . . . + \frac{1}{60} \left.\right)\)
\(S < \left(\right. \frac{1}{30} + \frac{1}{30} + . . . + \frac{1}{30} \left.\right) + \left(\right. \frac{1}{40} + \frac{1}{40} + . . . + \frac{1}{40} \left.\right) + \left(\right. \frac{1}{50} + \frac{1}{50} + . . . + \frac{1}{50} \left.\right)\)
\(S < \frac{10}{30} + \frac{10}{40} + \frac{10}{50}\)
\(S < \frac{47}{60} < \frac{50}{60} = \frac{5}{6}\)(1)
\(S > \left(\right. \frac{1}{40} + \frac{1}{40} + . . . + \frac{1}{40} \left.\right) + \left(\right. \frac{1}{50} + \frac{1}{50} + \frac{1}{50} + . . . + \frac{1}{50} \left.\right) + \left(\right. \frac{1}{60} + \frac{1}{60} + . . . + \frac{1}{60} \left.\right)\)
\(S > \frac{10}{40} + \frac{10}{50} + \frac{10}{60}\)
\(S > \frac{37}{60} > \frac{35}{60} \left(\right. 2 \left.\right)\)
Từ (1) và (2) => \(\frac{7}{12} < S < \frac{5}{6}\)
hay \(\frac{7}{12} < \frac{1}{31} + \frac{1}{32} + \frac{1}{33} + . . . + \frac{1}{59} + \frac{1}{60} < \frac{5}{6}\)