Question

Chứng minh rằng: \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)< 1

Answer 1

Ta có:

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+\dfrac{4}{4!}-\dfrac{1}{4!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)

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

Mà \(1-\dfrac{1}{100!}< 1\)

Vậy \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) (Đpcm)

Answer 2

\(\dfrac{1}{2!}\)+ \(\dfrac{2}{3!}\)+ \(\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\)

= \((\)\(\dfrac{1}{1!}\)-\(\dfrac{1}{2!}\)\()\) + \((\)\(\dfrac{1}{2!}\)-\(\dfrac{1}{3!}\)\()\) + \((\)\(\dfrac{1}{3!}\)-\(\dfrac{1}{4!}\)\()\) +...+ \((\)\(\dfrac{1}{99!}\)-\(\dfrac{1}{100!}\)\()\)

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