The Art and Craft of Problem Solving

320 CHAPTER 9 CALCULUS

To see why this is true, use the binomial theorem^3 to get

1+- = I +m- + -+ -+ -+ ...

( 1

)

m 1 (m) 1 (m) 1 (m) 1

m m 2 m^2 3 m^3 4 m^4

m(m- I) m(m-I)(m -2) m(m-l)(m-2) (m-3)

= I + I +

2'^2

+

3 '^3

+

4 '^4

+ ...

. m.m .m
I 1 I
< 1+1+-+-+-+···
2! 3! 4!
=e.

Consequently, (I + m) < em, so we have

S(x,n) = (I +x)(1 +�) ... (I +XZ)

< (/(/^2 ...

(/"

= (/+xZ+ ... +x".

Summing the geometric series, we conclude that

_^1 _ < S(x) < (/I(I-x),

I-x

for 0 < x < 1. Can these bounds be improved? •

Example 9.2.2 Fix a > I, and consider the sequence (xn)n?:O defined by Xo = a, and

Xn+ 1 = �

(

xn+ :), n=O,I,2, ....

Does this sequence converge, and if so, to what?

Solution: Let us try an example where a = (^5). Then we have
Xo =5,

XI = �

(

5 + �) = 3

X 2 = �

(

3 + �) = �.

Observe that the values (so far) are strictly decreasing. Will this always be the case?

Let us visualize the evolution of the sequence. If we draw the graphs of y = 5/ x and

y = x, we can construct a neat algorithm for producing the values of this sequence,

for Xn+ 1 is the average of the two numbers Xn and 5/xn. In the picture below, the y­

coordinates of points B and A are respectively Xo and 5/ Xo. Notice that the y-coordinate

of the midpoint of the line segment AB is the average of these two numbers, which is

equal to XI.

(^3) 0r use the fact that Iimm�""( I + I/m)m = e along with an analysis of the derivative of the function f(x) =
(I + Ilx)x to show that this limit is attained from below.