SEQUENCES       1 n 1 1 n 1 e lim 1 lim 1 1 n!1 n!1 n n n in .NET

Encoder QR Code in .NET SEQUENCES       1 n 1 1 n 1 e lim 1 lim 1 1 n!1 n!1 n n n

SEQUENCES       1 n 1 1 n 1 e lim 1 lim 1 1 n!1 n!1 n n n
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[CHAP. 2
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  1 x it follows that lim 1 e: x!1 x
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LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR 2.20. Find the (a) l.u.b., (b) g.l.b., 2; 2; 1; 1; 1; 1; 1; 1; . . . . (c) lim sup lim , and (d) lim inf (lim for the sequence
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(a) l:u:b: 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: 2, since all terms are greater than or equal to 2, while at least one term (the 2nd) is less than 2  for any  > 0. (c) lim sup or lim 1, since in nitely many terms of the sequence are greater than 1  for any  > 0 (namely, all 1 s in the sequence), while only a nite number of terms are greater than 1  for any  > 0 (namely, the 1st term).
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(d) lim inf or lim 1, since in nitely many terms of the sequence are less than 1  for any  > 0 (namely, all 1 s in the sequence), while only a nite number of terms are less than 1  for any  > 0 (namely the 2nd term).
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2.21. Find the (a) l.u.b., Problem 2.17.
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(b) g.l.b.,
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(c) lim sup (lim), and
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(d) lim inf (lim) for the sequences in
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The results are shown in the following table.
Sequence 2; 1:9; 1:8; 1:7; . . . ; 2 n 1 =10 . . . 1; 1; 1; 1; . . . ; 1
1 1 1 2; 3;4 n 1
l.u.b. 2 1
1 2 2 3
g.l.b. none 1 1 3 6 none
lim sup or lim 1 1 0
lim inf or lim 1 1 0
;... 1 ; . . .
n 1 1 = n 5 ; . . . ; 1 n
:6; :66; :666; . . . ; 2 1 3
1=10 ; . . .
1; 2; 3; 4; 5; . . . ; 1 n n; . . .
none
NESTED INTERVALS 2.22. Prove that to every set of nested intervals an ; bn , n 1; 2; 3; . . . ; there corresponds one and only one real number.
By de nition of nested intervals, an 1 A an ; bn 1 @ bn ; n 1; 2; 3; . . . and lim an bn 0. Then a1 @ an @ bn @ b1 , and the sequences fan g and fbn g are bounded and respectively monotonic increasing and decreasing sequences and so converge to a and b. To show that a b and thus prove the required result, we note that b a b bn bn an an a jb aj @ jb bn j jbn an j jan aj Now given any  > 0, we can nd N such that for all n > N jb bn j < =3; so that from (2), jb aj < . jbn an j < =3; jan aj < =3 3 Since  is any positive number, we must have b a 0 or a b. 1 2
CHAP. 2]
SEQUENCES
2.23. Prove the Bolzano Weierstrass theorem (see Page 6).
Suppose the given bounded in nite set is contained in the nite interval a; b . Divide this interval into two equal intervals. Then at least one of these, denoted by a1 ; b1 , contains in nitely many points. Dividing a1 ; b1 into two equal intervals, we obtain another interval, say, a2 ; b2 , containing in nitely many points. Continuing this process, we obtain a set of intervals an ; bn , n 1; 2; 3; . . . ; each interval contained in the preceding one and such that b1 a1 b a =2; b2 a2 b1 a1 =2 b a =22 ; . . . ; bn an b a =2n from which we see that lim bn an 0.
This set of nested intervals, by Problem 2.22, corresponds to a real number which represents a limit point and so proves the theorem.
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