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Unformatted text preview: e if condition a is false then so is
condition b. Suppose now that condition a is true; in other words, that sup A inf B. For every positive integer n, we
use the fact that
sup A 1
n
is not an upper bound of A to choose a member x n of A such that 168 sup A 1 xn.
n For every positive integer n we use the fact that
inf B 1
n
is not a lower bound of B to choose a member y n of B such that
y n inf B 1 .
n
Thus for each n we have
sup A 1 x n y n inf B 1
n
n
and therefore
0 yn xn 2
n
from which it follows that y n x n 0 as n Ý.
9. Suppose that S is a nonempty bounded set of real numbers. Prove that there exist two sequences x n and y n
in the set S such that
y n x n sup S inf S
as n Ý.
We have already seen that there exists a sequence x n in S such that x n sup S as n Ý. In the
same way we can see that there exists a sequence y n in S such that y n inf S as n Ý. We now
have
y n x n sup S inf S
as n Ý. Exercises on Monotone Sequences
1. Given that c 1, use the following method to prove that c n Ý: a. Write c 1, so that c 1 , and then use mathematical induction to prove that, if n is any positive
integer then c n 1 n.
For each positive integer n we take p n to be the assertion that c n 1 n. Since c 1 we
know that the assertion p 1 is true. Now suppose that n is any positive integer for which the
assertion p n is true. We observe that
1 1 n
c n1 cc n
1 n 1 n 2 1 n 1
and so the assertion p n1 is also true. We deduce from mathematical induction that the
inequality c n 1 n is true for every positive integer n.
b. Explain why 1 n Ý and then use this exercise in this subsection to show that c n Ý.
To prove that 1 n Ý as n Ý, suppose that w is a real number. The inequality
1 n w
will hold when
n w 1.
We choose a positive integer N such that
N w 1
and we observe that 1 n w whenever n N.
2. Given that c 1, use the following method to prove that c n Ý: a. Explain why the sequence c n is increasing and deduce that it has a limit. 169 The fact that the sequence c n is increasing follows at once from the inequality
c n1 cc n 1c n .
It now follows from the monotone sequences theorem that the sequence c n has a limit.
b. Call the limit x and show that if x is finite then the equation
c n1 cc n
leads to the equation x cx, which implies that x 0. But x cannot be equal to zero? Why not?
From the equation
c n1 cc n
we obtain
lim
lim c n1 n Ý cc n
nÝ
which gives us x cx. Since c n 1 for every n we know that x 1 and therefore x 0. 3. Suppose that c  1 and that, for every positive integer n,
n xn ci 1.
i1 Explain why
1 xn 1 c . From the identity
n 1 ci c 1 1 cn i1 we deduce that
n ci
i1 1 n
1 c
1c Now since 1/c  1 we know that 1/c 
Ý as n Ý and we conclude that c  n
Therefore c n 0 as n Ý and we conclude that
n
lim 1 c 1 .
nÝ 1
1c
c
n 0 as n Ý. 4. Suppose that x n is a given sequence of real numbers, that x 1 0 and that...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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