Download SEQUENCES, CONTINUED Definition 3.13. A sequence {sn} of real

SEQUENCES, CONTINUED
De…nition 3.13. A sequence fsn g of real numbers is said to be
1. increasing if sn+1
sn for n
1;
2. decreasing if sn+1
sn for n
1.
If fsn g is increasing or decreasing, we say that fsn g is monotonic.
Theorem 3.14. Let fsn g be a monotonic sequence. Then fsn g converges if and only if fsn g is
bounded.
Proof. See p. 55 of Rudin. We shall just prove the following statement:
If fsn g is increasing and bounded, then fsn g converges to sup E, where E = fsn : n 2 Ng is the
range of fsn g.
Let fsn g be an increasing and bounded sequence. Since E is bounded and nonempty, by the least
upper bound property we have that sup E 2 R exists. Since sup E is an upper bound for E, we have
sn sup E for all n 1. Since sup E is the least upper bound for E, for any " > 0, sup E " is not
an upper bound. Hence there exists N
1 such that sN > sup E ". Since fsn g is increasing, this
implies sn > sup E " for n N . So we have
sup E
" < sn
sup E
for n
N:
This implies jsn sup Ej < " for n N . We have proved that fsn g converges to sup E.
De…nition 3.15. We say that fsn g ! +1 (also written limn!1 sn = +1) if for each M 2 R
there exists N 2 N such that sn M for n N .
Similarly, we say that fsn g ! 1 (also written limn!1 sn = 1) if for each M 2 R there exists
N 2 N such that sn M for n N .
If fsn g converges, fsn g ! +1, or fsn g ! 1, then we say that fsn g converges to an extended
real number.
Remark. The set of extended real numbers is de…ned to be R = R [ f+1; 1g. The usual
order on R is extended to R by de…ning 1 < x and x < +1 for all x 2 R and 1 < +1.
Lemma. Let fsn g be a sequence of real numbers. Let S R be the set of extended real number
subsequential limits of fsn g. Then S is nonempty.
Proof. Case 1. fsn g is bounded. Then Theorem 3.6(b) implies that there exists a convergent
subsequence (converging to a real number).
Case 2. fsn g is unbounded. Then there a subsequence which converges either to +1 or to 1.
(Exercise: Prove this.)
Again, let S R denote the set of extended real number subsequential limits of a sequence fsn g.
De…ne
s = sup S;
s = inf S
to be the upper limit and lower limit of fsn g, respectively.
1
We use the notation:
s = lim sup sn ;
n!1
s = lim inf sn :
n!1
Recall that S R is the set of extended real number subsequential limits of fsn g.
Theorem 3.17.
(a) s 2 S and s 2 S.
(b) For each x > s there exists N 2 N such that sn < x for n N . For each x < s there exists
N 2 N such that sn > x for n N .
Moreover, s and s are the only numbers with properties (a) and (b).
Proof. See p. 56 of Rudin.
Example 3.18(a). Let fsn g be a sequence of real numbers whose range E contains all rational
numbers. In other words, for each q 2 Q there exists n 1 such that sn = q.
Exercise: (1) Prove that for each x 2 R, there exists a subsequence of fsn g converging to x.
Solution. Let x 2 R. Choose q1 2 Q such that q1 2 (x; x + 1). Since fsn g contains all rational
numbers, there exists n1 2 N such that q1 = sn1 . Now assume that we have chosen n1 ; : : : ; nk 1 2 N.
Choose qk 2 Q such that qk 2 (x; x + k1 ). Since fsn g contains all rational numbers and since the
interval (x; x + k1 ) contains an in…nite number of rationals, there exists nk > nk 1 such that qk = snk .
Then n1 < n2 <
and jsnk xj < k1 for each k 1. Hence the subsequence fsnk g converges to x.
(2) Prove that lim supn!1 sn = +1 and (similarly) lim inf n!1 sn = 1.
Theorem 3.20(c).
p
lim n n = 1:
n!1
Proof. See p. 58 of Rudin.
Let e be Euler’s number.
Theorem 3.31.
lim
n!1
1+
Proof. See p. 64 of Rudin.
P
Here e is de…ned by the series sum e = 1
n=0
1
n
1
;
n!
2
n
= e:
(in…nite) series is a topic we consider next.