Page 41 - Schaum's Outline of Theory and Problems of Advanced Calculus

32 SEQUENCES [CHAP. 2

nþ1
1 1 1
n
and lim 1 þ ¼ lim 1 þ 1 þ ¼ e
n n n
n!1 n!1
x
1
¼ e:
x
it follows that lim 1 þ
x!1
LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR
2.20. Find the (a) l.u.b., (b) g.l.b., (c) lim sup ðlimÞ, and (d) lim inf (limÞ for the sequence
2; 2; 1; 1; 1; 1; 1; 1; ... .
(a)l:u:b: ¼ 2, since all terms are less than equal to 2, while at least one term (the 1st) is greater than 2
for any > 0.
(b)g:l:b: ¼ 2, since all terms are greater than or equal to 2, while at least one term (the 2nd) is less than
2 þ for any > 0.
(c) lim sup or lim ¼ 1, since infinitely many terms of the sequence are greater than 1 for any > 0
(namely, all 1’s in the sequence), while only a finite number of terms are greater than 1 þ for any > 0
(namely, the 1st term).
(d) lim inf or lim ¼ 1, since infinitely many terms of the sequence are less than 1 þ for any > 0
(namely, all 1’s in the sequence), while only a finite number of terms are less than 1 for any > 0
(namely the 2nd term).

2.21. Find the (a) l.u.b., (b) g.l.b., (c) lim sup (lim), and (d) lim inf (lim) for the sequences in
Problem 2.17.
The results are shown in the following table.

Sequence l.u.b. g.l.b. lim sup or lim lim inf or lim
2; 1:9; 1:8; 1:7; ... ; 2 ðn 1Þ=10 .. . 2 none 1 1
n 1
1; 1; 1; 1; ... ; ð 1Þ ; ... 1 1 1 1
1 1 1 1 n 1 =ðn þ 1Þ; ... 1 1 0 0
5
2 ; ; ; .. . ; ð 1Þ 2 3
3 4
n
2
:6;:66;:666; .. . ; ð1 1=10 Þ; ... 2 6 2 2
3 3 3 3
n
1; þ2; 3; þ4; 5; .. . ; ð 1Þ n; .. . none none þ1 1
NESTED INTERVALS
2.22. Prove that to every set of nested intervals ½a n ; b n Š, n ¼ 1; 2; 3; .. . ; there corresponds one and only
one real number.
By definition of nested intervals, a nþ1 A a n ; b nþ1 @ b n ; n ¼ 1; 2; 3; .. . and lim ða n b n Þ¼ 0.
n!1
Then a 1 @ a n @ b n @ b 1 , and the sequences fa n g and fb n g are bounded and respectively monotonic
increasing and decreasing sequences and so converge to a and b.
To show that a ¼ b and thus prove the required result, we note that
b a ¼ðb b n Þþðb n a n Þþða n aÞ ð1Þ
jb aj @ jb b n jþjb n a n jþja n aj ð2Þ
Now given any > 0, we can find N such that for all n > N
jb b n j < =3; jb n a n j < =3; ja n aj < =3 ð3Þ
so that from (2), jb aj < . Since is any positive number, we must have b a ¼ 0or a ¼ b.