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\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
A = (\(\dfrac{1}{2}\) - 1)(\(\dfrac{1}{3}\)-1)(\(\dfrac{1}{4}\)-1)......(\(\dfrac{1}{2018}\)-1)(\(\dfrac{1}{2019}\)-1)
A = \(\dfrac{-1}{2}\).\(\dfrac{-2}{3}\).\(\dfrac{-3}{4}\).........\(\dfrac{-2017}{2018}\).\(\dfrac{-2018}{2019}\)
xét dãy số -1; -2; -3;.....-2017; -2018 có 2018 số hạng
vậy A = \(\dfrac{1}{2019}\)
B = \(\dfrac{-1}{2}\).\(\dfrac{-2}{3}\).\(\dfrac{-3}{4}\)..................\(\dfrac{-2017}{2018}\).\(\dfrac{-2018}{2019}\)
xét dãy số -1; -2; -3; -4;....-2017; -2018
số số hạng của dãy số trên là {-2018 -(-1)}: (-1)+1 = 2018
vậy B = \(\dfrac{1}{2019}\)