Question

Chứng minh:

\(\frac{1}{2!}+\frac{1}{3!}+.....+\frac{1}{2016!}<1\)

Answer 1

đặt tổng trên là A ta có

A<1/1.2+1/2.3+1/3.4+.....+1/2015.2016

    =1-1/2016<1

=>A<1 (đpcm)

Answer 2

Đặt A= \(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2016!}\)

Có:\(\frac{1}{2!}\)<\(\frac{1}{1.2}\)

\(\frac{1}{3!}<\frac{1}{2.3}\)

\(\frac{1}{4!}<\frac{1}{3.4}\)

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

\(\frac{1}{2016!}<\frac{1}{2015.2016}\)

=> A <\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2015.2016}\)

=> A<\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

=> A<1-\(\frac{1}{2016}\)<1

=> A<1

( T I C K CHO MÌNH NHA )