Problem: (2000 Putnam – A1) Let be a positive real number. What are the possible values of , given that are positive numbers such that ?
Solution: First, we will try to guess what the range is using inequalities. We obviously have
because all the are positive. Also,
by “expansion” (infinitely, but that’s ok). So for now, our guess is the interval . We want to show that given any we can make a sequence such that and . Well, we like easy sequences so consider if is a geometric sequence with common ratio . Then
We want to show that given any , we can find a positive real satisfying , the above equation, and
Rewriting the two equations (and squaring the first), we have
Since we want , we must use the second factor, which gives us
But if , clearly , so there exists some positive ratio such that and . Hence the possible values are . QED.
Comment: A surprisingly difficult A1 problem on a Putnam; it had less perfect scores than A2, B1, and B2 (which will probably be posted over the next week or so). It required some experience with inequalities to guess the range and then a clever assumption of a geometric series to complete the proof.
Practice Problem: What if the terms could be any nonzero real? What are the possible values of then?
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