University of South Carolina High School Math Contest/1993 Exam/Problem 22

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Problem

Let

$A = \left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 + \frac 13 + \frac 19\right) \left( 1 + \frac 15...$ $B = \left( 1 - \frac 12\right)^{-1} \left( 1 - \frac 13 \right)^{-1} \left(1 - \frac 15\right)^{-1} \left(1 - \frac 17\right)...$

and

$C = 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \frac 19 + \frac 1{10} + \frac 1{11} + \...$

Then which of the following inequalities is true?

$\mathrm{(A) \ } A > B > C \qquad \mathrm{(B) \ } B > A > C \qquad \mathrm{(C) \ } C > B > A \qquad \mathrm...$

Solution

If you imagine expanding out the expression for $A$ , you can see that every term in $C$ will appear once, along with plenty of others. (Think of the prime factorizations and you can figure out which products give the terms of $C$ .) Since all terms are positive, $A > C$ .