Question

B = \(\dfrac{9}{10!}+\dfrac{10}{11!}+....+\dfrac{99}{100!}\)

Chứng minh B < \(\dfrac{1}{9}\)

Answer 1

\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)

Ta thấy :

\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)

\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)

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

\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)

\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)