Đặt:
`A = 1/(1.101) + 1/(2.102) + 1/(3.103) + ... + 1/(10.110)`
`100A = 100/(1.101) + 100/(2 . 102) + 100/(3.103) + ... + 10/(10.110)`
`100A = 1 - 1/101 + 1/2 - 1/102 + 1/3 - 1/103 + ... + 1/10 - 1/110`
`100A = (1 + 1/2 + 1/3 + ... + 1/10) - (1/101 + 1/102 + 1/103 + .... + 1/110)`
$\\$
Đặt:
`B = 1/(1.11) + 1/(2.12) + 1/(3.13) + ... + 1/(100.110)`
`10B = 10/(1.11) + 10/(2.12) + 10/(3.13) + ... + 10/(100.110)`
`10B = 1 - 1/11 + 1/2 - 1/12 + 1/3 - 1/13 + ... + 1/100 - 1/110`
`10B = (1 + 1/2 + 1/3 + ... + 1/100) - (1/11 + 1/12 + 1/13 + ... + 1/110)`
Do `(1 + 1/2 + 1/3 + ... + 1/100) - (1/11 + 1/12 + 1/13 + ... + 1/110) = (1 + 1/2 + 1/3 + ... + 1/10) - (1/101 + 1/102 + 1/103 + .... + 1/110)`
`=> 100A = 10B`
`=> A/B = 10/100`
`=> A/B = 1/10`
Do đó:
$\dfrac{\dfrac{1}{1.100} + \dfrac{1}{2.102} + \dfrac{1}{3.103} + ... + \dfrac{1}{10 . 110}}{\dfrac{1}{1.11} + \dfrac{1}{2.12} + \dfrac{1}{3.13} + ... + \dfrac{1}{100.110}}$
`= 1/10`
Vậy tỉ số của $\dfrac{\dfrac{1}{1.100} + \dfrac{1}{2.102} + \dfrac{1}{3.103} + ... + \dfrac{1}{10 . 110}}{\dfrac{1}{1.11} + \dfrac{1}{2.12} + \dfrac{1}{3.13} + ... + \dfrac{1}{100.110}}$ là `1/10`
