helpppppppppppppppppp

Với `n in NN**` , ta có :

`S = 1/3 + 3/(3.7) + 7/(3.7.11.15) + ...... + (2n+1)/(3.7.11.....(4n+3))` ( với `n in NN**`)

`S = (2.0+1)/(4.0+3) + (2.1+1)/(3.(4.1+3)) + (2.2+1)/(3.7.(4.2+3)) + (2.3+1)/(3.7.11.(4.3+3)) + ......... + (2n+1)/(3.7.11.15.....(4n+3))`

`S = 1/2 . [ (2.(2.0+1))/(4.0+3) + (2.(2.1+1))/(3.(4.1+3)) + (2.(2.2+1))/(3.7.(4.2+3)) + (2.(2.3+1))/(3.7.11.(4.3+3)) + ......... + (2.(2n+1))/(3.7.11.15.....(4n+3)) ]`

`S = 1/2 . [ (4.0+2)/(4.0+3) + (4.1+2)/(3.(4.1+3)) + (4.2+2)/(3.7.(4.2+3)) + (4.3+2)/(3.7.11.(4.3+3)) + ......... + (4n+2)/(3.7.11.15.....(4n+3)) ]`

`S = 1/2 . [ ((4.0+3)-1)/(4.0+3) + ((4.1+3)-1)/(3.(4.1+3)) + ((4.2+3)-1)/(3.7.(4.2+3)) + ((4.3+3)-1)/(3.7.11.(4.3+3)) + ......... + ((4n+3)-1)/(3.7.11.15.....(4n+3)) ]`

`S = 1/2 . [ (-1)/(4.0+3) + (1/3 - 1/(3.(4.1+3))) + (1/(3.7) - 1/(3.7.(4.2+3))) + (1/(3.7.11) - 1/(3.7.11.(4.3+3))) + ...... + (1/(3.7.11.15....(4.(n-1)+3))) ]` 

`S = 1/2. [ ((-1)/3 + 1/3) + ((-1)/(3.7) + 1/(3.7)) + ((-1)/(3.7.11) + 1/(3.7.11)) + ((-1)/(3.7.11.15) + 1/(3.7.11.15)) + ...... + ((-1)/(3.7.11.15.....(4.(n-1)+3)) + 1/(3.7.11.15.....(4.(n-1)+3))) ]`

`S = 1/2 . 0 = 0`

Mà `0 < 1/2`

`=> S < 1/2` 

Vậy `S < 1/2`