Question

Chứng minh rằng :

A=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+....+\frac{2!}{n!}< 1\)

Answer 1

\(A=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+....+\frac{2!}{n!}\)

\(A=2!.\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+....+\frac{1}{n!}\right)\)

\(A< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{\left(n-1\right).n}\right)\)

\(A< 2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(A< 2.\left(\frac{1}{2}-\frac{1}{n}\right)\)

\(A< 1-\frac{2}{n}< 1\)

=> \(A< 1\)