`B = 1/3 - 1/(3^2) + 1/(3^3) + 1/(3^4) + ... + 1/(3^99) - 1/(3^100)`

`3B = 1-1/3+1/(3^2) + 1/(3^3)+ ... + 1/(3^100) - 1/(3^101)`

`3B - B = (1-1/3+1/(3^2) + 1/(3^3)+ ... + 1/(3^100) - 1/(3^101))-(1/3 - 1/(3^2) + 1/(3^3) + 1/(3^4) + ... + 1/(3^99) - 1/(3^100))`

`2B = 1-2/3+2/(3^100) - 1/(3^101)`

`B = (1/3 +2/(3^100) - 1/(3^101))/2`

`------------`

`(5^20+1)/(5^20) = 1 + 1/(5^20)`

`(5^21+1)/(5^21) = 1 + 1/(5^21)`

Vì `1/(5^20) > 1/(5^21)`

nên `(5^20+1)/(5^20) > (5^21+1)/(5^21)`