Lời giải:
Ta có:
\(\frac{1}{5!}=\frac{1}{1.2.3.4.5}< \frac{1}{3.4.5}\)
\(\frac{1}{6!}< \frac{1}{4.5.6}\)
.........
\(\frac{1}{2019!}< \frac{1}{2017.2018.2019}\)
Do đó:
\(C< 1+\frac{1}{2}+\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2017.2018.2019}\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2019-2017}{2017.2018.2019}\right)\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)< \frac{3}{2}+\frac{1}{2}.\frac{1}{1.2}\)
\(C< \frac{7}{4}\)
