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