94 Chapter 2 • Sequences
Thus, by Theorem 2.3.5, lim O.d 1 d 2 · · · dn = 1. Since a sequence cannot
n-+oo
have more then one limit, 0.999 · · · 9 · · · == 1.000 · · · 0 · · ·. 0
We summarize this discussion (including the case x < 0) in the following
theorem.
*Theorem 2.5.7 Every real number x can be represented as an (infinite) dec-
imal expansion, x = K.d 1 d2 · · · dndn+l · · ·, where dn E {O, 1, 2, 3, · · · , 9}. This
decimal representation is unique except when one of them ends in all O's and
the other in all 9 's.MORE APPLICATIONSExample 2.5.8 Consider the sequence {xn} defined inductively by x 1 = 1,
and Vn EN, Xn+i = Jxn + 2. P rove that {xn} converges, and find its limit.
Solution: Part 1- Prove that {xn} converges.(a) First, we prove that {xn } is monotone increasing. We use mathematical
induction to prove the following: Vn EN, 0:::; Xn :::; Xn+l·
(i) X1 = 1 and X2 = Jf+2 = V3. Thus, 0:::; X1 :::; X2.
(ii) Now, assume 0 :::; Xk :::; Xk+l · Then ,
0 :::; Xk + 2 :::; Xk+l + 2
0 :::; VXk + 2 :::; VXk+l + 2
0 :::; Xk+l < Xk+2·
Therefore, by (i), (ii), and mathematical induction, Vn EN, 0:::; Xn:::; Xn+l ·
That is, { Xn} is monotone increasing.
(b) Now, we prove that {xn} is bounded above. We use mathematical
induction to prove the following: Vn E N, Xn :::; 3.
(i) X1 = 1 SO X1 :::; 3.
(ii) Now, assume Xk :::; 3. Then
Xk + 2 :::; 5
VXk + 2 :::; vis
Xk+l :::; 3.
Therefore, by (i), (ii), and mathematical induction, Vn E N, Xn :::; 3. That
is, {xn} is bounded above by 3.
(c) Therefore, by (a), (b), and t he monotone convergence theorem, {xn}
converges.