The Art and Craft of Problem Solving

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9.2 CONVERGENCE AND CONTINUITY 321

y=x

2 � � 4 • 6 8

y=5/x

Next, draw a horizontal line left from this midpoint until it intersects the graph of
y = x (at C). The coordinates of C are (Xl ,Xl), and we can drop a vertical line from C
until it meets the graph of y = 51 x (at D). By the same reasoning as before, X 2 is the
y-coordinate of the midpoint of segment CD.
Continuing this process, we reach the point E = (X 2 ,X 2 ), and it seems clear from
this picture that if we keep going, we will converge to the intersection of the two
graphs, which is the point (v's, v's).
Thus we conjecture that limn-.ooxn = v's. However, the picture is not a rigorous
proof, but an aid to reasoning. To show convergence with this picture, we would
need to argue carefully why we will never "bounce away" from the convergence point.
While it is possible to rigorize this, let's change gears and analyze the general problem
algebraically.
The picture suggests two things: that the sequence decreases monotonically, and
that it decreases to Va. To prove monotonicity, we must show that Xn+l :S Xn• This is
easy to do by computing the difference

(x�+a) 2x�-x�-a _�-a

Xn Xn+ l -Xn

2xn

• 2xn - 2xn '
  which is non-negative as long as x� 2 a. And this last inequality is true; it is a simple
  consequence of the AM-OM inequality (see page 17 6):

so Xn+ l 2 Va no matter what Xn is equal to.^4 Since Xo = a > Va, all terms of the

sequence are greater than or equal to Va.

4 Instead of studying the difference Xn -Xn+ 1, it is just as easy to look at the ratio xn+ 1/ Xn• This is always less
than or equal to I (using a little algebra and the fact that Xn � Va).