Question

[Hồng Diễm Nguyễn]

1+1/2^2+1/3^2+.......+1/100^2

Answer 1

[Nguyễn Hữu Huân]

Ta có:

$\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{{100}^2}<\frac{1}{99.100}$

Đặt:

$A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}$

$\to A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=2-\frac{1}{100}<2$

$\to 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}<2$

$\to$đpcm

nhớ tick cho mik nha

Answer 2

[Nguyễn Hữu Huân]

cách 2 : 

Đặt 1+1/2^2+1/3^2+...+1/100^2=A

Có A<1+1/1.2+1/2.3+1/3.4+....+1/99.100

=>A<1+1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100

=>A<1+1-1/100

=>A<2-1/100<2

nhớ tickkk