Question

Chứng tỏ A<1 biết :

A=\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{2013}{2014!}\)

Answer 1

Ta có:

\(A=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{2013}{2014!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{2014-1}{2014!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+...+\dfrac{2014}{2014!}-\dfrac{1}{2014!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{2013!}-\dfrac{1}{2014!}\)

\(=\dfrac{1}{1!}-\dfrac{1}{2014!}=1-\dfrac{1}{2014!}\)

Do \(1-\dfrac{1}{2014!}< 1\) Nên \(A< 1\)

Vậy \(A=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{2013}{2014!}< 1\) (Đpcm)

Answer 2

tick nhé