### 7th Putnam 1947

Problem A1

The sequence an of real numbers satisfies an+1 = 1/(2 - an). Show that limn→∞an = 1.

Solution

Fairly easy.

This is slightly messy. First, since k = 1/(2 - k) implies k = 1, it is obvious that if the sequence tends to a limit, then the limit is 1.

Next, if 0 < an < 1, then 1 < 2 - an < 2, so 1/2 < an+1 < 1. So once the sequence gets into the interval (0, 1) it stays there. But an+1 - an = (an - 1)2/(2 - an) > 0 for an < 2   (*). So once the sequence gets into the interval (0, 1) it is monotonic increasing and bounded above by 1, and hence tends to a limit (which must be 1).

If an < 0, then 0 < an+1 < 1, so we are also home if a member of the sequence is negative. Similarly, an > 2 implies an+1 < 0. If an = 1, then all following terms are 1 and so the limit is 1. So the only issue is if 1 < a1 < 2.

But then (*) shows that whilst the sequence remains in the interval (1, 2) it is monotonic increasing. It cannot tend to a limit, because that limit would have to be 1, which is impossible. So it cannot stay in the interval. We cannot have am = 2, because then we would not have am+1 = 1/(2 - am), so we must have am > 2 for some m and then we are home.

The official solutions have a neater solution:

Let bn = 1/(1 - an). Then bn+1 = (2 - an)/(1 - an) = bn + 1. Hence for some β, bn = n + β, and hence an = (n + β - 1)/(n + β). With a little tidying up, this is a solution.