cmr 1-1/2+1/3-1/4+...+1/199-1/200<5/6
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Xét vế phải\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}\)
=\(\left(1+\frac{1}{3}+\frac{1}{5}+..+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-...-\frac{1}{100}\)
=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)
\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)
\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)
\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)
\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)
\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{199}+\dfrac{1}{200}\\ >\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)\left(90so\right)+\left(\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\right)\left(100so\right)\\ A>\dfrac{1}{10}+\dfrac{90}{100}+\dfrac{100}{200}=1+\dfrac{1}{2}=\dfrac{3}{2}\left(đpcm\right).\)