We have 0 < a < b which means 1/a > 1/b and so 3 + 1/a > 3 + 1/b. Flipping over the fraction, we will get 1/(3 + 1/a) < 1/(3 + 1/b) and the inequality remains the same way round when 2 is added. Flipping over again for the last time we get

1 2+ 1 3+ 1 a

is greater than

1 2+ 1 3+ 1 b

The second part is a further expansion of the first, and in the process of repeating the above we know that it involves just one more flipping over of the fraction, thus

1 2+ 1 3+ 1 4 + [1/(a)]

is less than the same thing with b in place of a as the inequality would be reversed again.

Lastly the continued fractions are expanded all the way down to 100 + 1/a and 100 + 1/b. Observe the above process, we can tell that if the last or biggest number is odd then the continued fraction with a in it is bigger. If the last or biggest number is even then the continued fraction with b in it is bigger. Each successive continued fraction involves one more 'flipping over' and reverses the inequality one more time. The following continued fraction is smaller than the same thing with b in place of a:

1 2 + 1 3+ 1 4 + ... + 1 99+ 1 100 + [1/( a)]