Theo đầu bài ta có:
\(\frac{3}{5\cdot2!}+\frac{3}{5\cdot3!}+\frac{3}{5\cdot4!}+...+\frac{3}{5.100!}< 0,6\)
\(\Rightarrow\frac{3}{5}\cdot\frac{1}{2!}+\frac{3}{5}\cdot\frac{1}{3!}+\frac{3}{5}\cdot\frac{1}{4!}+...+\frac{3}{5}\cdot\frac{1}{100!}< \frac{3}{5}\)
\(\Rightarrow\frac{3}{5}\cdot\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)< \frac{3}{5}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1\)( điều cần chứng minh )
Mà \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1-\frac{1}{100}< 1\)( đã chứng minh được )
Vậy \(\frac{3}{5\cdot2!}+\frac{3}{5\cdot3!}+\frac{3}{5\cdot4!}+...+\frac{3}{5\cdot100!}< 0,6\)( đpcm )
