Ta có: \(\frac{1}{5!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}< \frac{1}{3\cdot4\cdot5}\)

\(\frac{1}{6!}< \frac{1}{1\cdot2\cdot3\cdot4\cdot5\cdot6}< \frac{1}{4\cdot5\cdot6}\)

..............

\(\frac{1}{2019!}=\frac{1}{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2019}< \frac{1}{2017\cdot2018\cdot209}\)

Do đó 

\(C< 1+\frac{1}{2}+\frac{1}{2\cdot3\cdot4}+\frac{1}{4\cdot5\cdot6}+....+\frac{1}{2017\cdot2018\cdot2019}\)

\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+.....+\frac{2019-2017}{2017\cdot2018\cdot2019}\right)\)

\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2018\cdot2019}\right)< \frac{3}{2}+\frac{1}{2}\cdot\frac{1}{1\cdot2}\)

\(\Rightarrow C< \frac{7}{4}\)

Nguồn: Nock Nock

\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)

\(=\frac{1}{1}+\frac{1}{1.2}+\frac{1}{1.2.3}+...+\frac{1}{1.2.3...2019}\)

\(=\frac{1}{1}+\frac{1}{1}.\frac{1}{2}+\frac{1}{1}.\frac{1}{2}.\frac{1}{3}+...+\left(\frac{1}{1}.\frac{1}{2}.\frac{1}{3}...\frac{1}{2018}.\frac{1}{2019}\right)\)

\(=\left(1.1.1....1.1\right)+\left(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}\right)+\left(\frac{1}{3}.\frac{1}{3}.\frac{1}{3}...\frac{1}{3}.\frac{1}{3}\right)+...+\left(\frac{1}{2018}.\frac{1}{2018}\right)+\frac{1}{2019}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}\)

Nhận xét rằng:

\(1< \frac{7}{8076};2< \frac{7}{8076};3< \frac{7}{8076};...;\frac{1}{1154}>\frac{7}{8076};\frac{1}{1155}>\frac{7}{8076};...;\frac{1}{2018}>\frac{7}{8076};\frac{1}{2019}>\frac{7}{8076}\)

Do đó:

\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}>\frac{7}{8076}+\frac{7}{8076}+...+\frac{7}{8076}\)

Vì tổng C có 2019 số hạng, suy ra \(C>2019.\frac{7}{8076}=\frac{7}{4}\)

Mình nhầm một chút:

\(1>\frac{7}{8076};\frac{1}{2}>\frac{7}{8076};\frac{1}{3}>\frac{7}{8076};...;\frac{1}{1154}< \frac{7}{8076};\frac{1}{1155}< \frac{7}{8076};...;\frac{1}{2019}< \frac{7}{8076}.\)

Do phân số lớn hơn chiếm phần nhiều nên:

\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}>\frac{7}{8076}+\frac{7}{8076}+...+\frac{7}{8076}\)

\(\Rightarrow C>2019.\frac{7}{8076}=\frac{7}{4}\)

Tham khảo nhé

Câu hỏi của Assassin_07 - Toán lớp 7 - Học toán với OnlineMath

Nguyễn Trần Nhật Anh , đâu có cầnnn

B = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...+\frac{1}{1+2+3+...+2019}\)

    = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2019\times1010}\)

    = \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2019\times2020}\right)\)

   = \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2019\times2020}\right)\)

  = \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

  = \(2\times\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(=2\times\frac{1009}{2020}\)

\(=\frac{1009}{1010}< \frac{1010}{1010}=1\)

\(\Rightarrow B< 1\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^

nhầm dấu rồi bé hơn 1 chứ:v bạn sửa lại hộ mình nhà