`a)` Ta có :
`1/2^2 = 1/2.2 < 1/1.2`
`1/3^2 = 1/3.3 < 1/2.3`
`1/4^2 = 1/4.4 < 1/3.4`
`....`
`1/45^2 = 1/45.45 < 1/ 44.45`
`=> 1/2^2+1/3^2+1/4^2+...+1/45^2 < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/44.45`
`=> 1/2^2+1/3^2+1/4^2+...+1/45^2<1 - 1/2 + 1/2 - 1/3 + 1/3- 1/4 + ... + 1/44 -1/45`
`=> 1/2^2+1/3^2+1/4^2+...+1/45^2<1 -1/45`
`=> 1/2^2+1/3^2+1/4^2+...+1/45^2<1 `
`b) 1/5+1/6+1/7+.... + 1/19`
` = (1/5 + 1/6 +1/7 +1/8 + 1/9) + (1/10 + 1/11 + 1/12 + .... + 1/19)`
Ta thấy :
`1/5 = 1/5`
`1/6 < 1/5`
`1/7 < 1/5`
`1/8 <1/5`
`1/9 < 1/5`
`=> 1/5 + 1/6 +1/7 +1/8 + 1/9 < 1/5 + 1/5 + 1/5 + 1/5 + 1/5`
`=> 1/5 + 1/6 +1/7 +1/8 + 1/9 < 5 . 1/5`
`=> 1/5 + 1/6 +1/7 +1/8 + 1/9 < 1 (1)`
Ta thấy :
`1/10 = 1/10`
`1/11 < 1/10`
`1/12 <1/10`
`.....`
`1/19 < 1/10`
`=> 1/10 + 1/11 + 1/12 + .... + 1/19 < \underbrace{1/10 +1/10 + 1/10 + ...+1/10}`
có `10` số hạng
`=> 1/10 + 1/11 + 1/12 + .... + 1/19 < 1/10 . 10`
`=> 1/10 + 1/11 + 1/12 + .... + 1/19 < 1 (2)`
Từ `(1)` và `(2)` suy ra :
`(1/5 + 1/6 +1/7 +1/8 + 1/9 ) + (1/10 + 1/11 + 1/12 + .... + 1/19) < 1 + 1`
`=> 1/5+1/6+1/7+.... + 1/19<2 `
