Question

chứng minh rằng:\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}<1\)

Answer 1

Ta có đẳng thức sau :  \(\frac{n-1}{n!}=\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)

Áp dụng đẳng thức trên được : 

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

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

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

Vậy ta có điều phải chứng minh.

Answer 2

Cho n\(\in\)N*.CMR:

\(\frac{n-1}{n!}=\frac{1}{n-1!}-\frac{1}{n!}\)

Ta có:\(\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{n-1!}-\frac{1}{n!}\)

Từ đó suy ra:

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

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

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

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

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

Suy ra:\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.........+\frac{99}{100!}< 1\)

Answer 3

1/1!-1/2!+1/2!-1/3!+.....+1/99!-1/100!

= 1-1/100! < 1 (đpcm)