Question

Chứng minh rằng: 1/2!+2/3!+3/4!+...+99/100!<1

Answer 1

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)

\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

\(=1-\frac{1}{100!}< 1\)

Answer 2

=> 1/1-1/2+1/2-1/3+1/3...1/99+1/99-1/100

=>1/1-(0)-1/100

1/1-1.100

=99/100

=>1/2+2/3,...<1