Chứng minh rằng a) 1/2^2+1/3^2+1/4^2+…+1/45^2<1 b) 1/5+1/6+1/7+....1/19<2

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1. `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 `

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