Question

[Cuber Việt]

Cho A gồm 100 số hạng :\(\dfrac{1}{1.1!}\) + \(\dfrac{1}{2.2!}\) + \(\dfrac{1}{3.3!}\) + ... + \(\dfrac{1}{2013.2013!}\)

Chứng minh rằng : A < \(\dfrac{3}{2}\)

Answer 1

[Nguyễn Huy Tú]

@Ace Legona

Answer 2

[Mashiro Shiina]

Nhận thấy \(\)\(\dfrac{1}{1.1!}=1\); \(\dfrac{1}{2.2!}=\dfrac{1}{4}\)

Đặt \(P=\dfrac{1}{3.3!}+...+\dfrac{1}{2013.2013!}\)

\(P=\dfrac{1}{3.1.2.3}+...+\dfrac{1}{2013.1.2...2013}\)

\(P< \dfrac{1}{1.2.3}+...+\dfrac{1}{2011.2012.2013}\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}-\dfrac{1}{2012.2013}\right)\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2012.2013}\right)=\dfrac{1}{4}-\dfrac{1}{2.2012.2013}\)

\(P< \dfrac{1}{4}\)

\(A< \dfrac{1}{4}+\dfrac{1}{4}+1=\dfrac{3}{2}\left(đpcm\right)\)