Question

[Nguyễn Thành Đạt]

Chứng Minh Rằng

1/2! + 2/3! + 3/4!+...+n/(n+1)! < 1

Answer 1

[hnamyuh]

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{\left(n+1\right)-1}{\left(n+1\right)!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+...+\dfrac{\left(n+1\right)}{\left(n+1\right)!}-\dfrac{1}{\left(n+1\right)!}\)

\(=1-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}-\dfrac{1}{\left(n+1\right)!}\)

( Vì \(\dfrac{3}{3!}=\dfrac{1}{2!};\dfrac{4}{4!}=\dfrac{1}{3!};...;\dfrac{n+1}{\left(n+1\right)!}=\dfrac{1}{n!}\))

\(=1-\dfrac{1}{\left(n+1\right)!}< 1\)

Answer 2

[Lê Song Phương]

Đặt \(S\left(n\right)=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)

Ta có \(S\left(1\right)=\dfrac{1}{2!}=\dfrac{1}{2}=1-\dfrac{1}{2!}\)

\(S\left(2\right)=S\left(1\right)+\dfrac{2}{3!}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}=1-\dfrac{1}{3!}\)

\(S\left(3\right)=S\left(2\right)+\dfrac{3}{4!}=\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{23}{24}=1-\dfrac{1}{4!}\)

Từ đây, ta có \(S\left(n\right)=1-\dfrac{1}{\left(n+1\right)!}\) và hiển nhiên \(S\left(n\right)< 1\) do \(\dfrac{1}{\left(n+1\right)!}>0\)

Vậy ta có đpcm