Theorem on Limits of Bounded Monotone Increasing Sequences, Study notes of Mathematics

A proof of the theorem stating that every bounded monotone increasing sequence has a limit. The proof uses the binary expansion of numbers and the bin property. Students of mathematics, particularly those studying sequences and series, will find this document useful.

Typology: Study notes

Pre 2010

Uploaded on 07/29/2009

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MthT 430 Notes Chap7c Bounded Monotone Sequences Have Limits
BISHL: Bounded Increasing Sequences Have Limits
Theorem. Let {xn}
n=1 be a bounded monotone increasing sequence; i.e.
x1x2. . . ,
and there is a number Msuch that for n= 1,2, . . .,
xnM.
Then there is a number Lsuch that
lim
n→∞
xn=L.
Proof using (P13–BIN): Without loss of generality, we assume that
0x1x2... xn. . . < 1.
We will construct a binary expansion for L.
A picture is helpful!
Divide the interval [0,1) into two halves.
Is there an n1such that xn1m0=1
2= 0.bin1?
If NO, let c1= 0, a1= 0 = 0.binc1,b1=m0=a1+1
2. If YES, let c1= 1,
a1=m0=1
2= 0.binc1,b1=a1+1
2= 1.
In both cases, for nn1,a1= 0.binc1xnb1=a1+1
21and b1a1=1
21.
Next Divide the interval [a1,b 1) = ·0.binc1, a1+1
21into two halves.
Is there an n2> n1such that xn2m1=a1+1
22?
If NO, let c2= 0, a2=a1= 0.binc1c2,b2=m1=a2+1
22. If YES, let c2= 1,
a2=m1= 0.binc1c2,b2=b1=a2+1
22.
In both cases, for nn2,a2= 0.binc1c2xn< b2=a2+1
22and b2a2=1
22.
chap7c.pdf page 1/2
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MthT 430 Notes Chap7c Bounded Monotone Sequences Have Limits

BISHL: Bounded Increasing Sequences Have Limits

Theorem. Let {xn}

n=

be a bounded monotone increasing sequence; i.e.

x 1 ≤ x 2

and there is a number M such that for n = 1, 2 ,.. .,

x n

≤ M.

Then there is a number L such that

lim

n→∞

xn = L.

Proof using (P13–BIN): Without loss of generality, we assume that

0 ≤ x 1 ≤ x 2 ≤... ≤ x n

We will construct a binary expansion for L.

A picture is helpful!

Divide the interval [0, 1) into two halves.

Is there an n 1 such that x n 1 ≥ m 0

1

2

bin

If NO, let c 1 = 0, a 1 = 0 = 0. bin

c 1 , b 1 = m 0 = a 1 +

1

2

. If YES, let c 1 = 1,

a 1 = m 0

1

2

bin

c 1 , b 1 = a 1

1

2

In both cases, for n ≥ n 1 , a 1

bin

c 1 ≤ x n ≤ b 1 = a 1

1

and b 1 − a 1

1

Next Divide the interval [a 1 , b 1

[

bin

c 1 , a 1

1

into two halves.

Is there an n 2

n 1 such that x n 2 ≥ m 1 = a 1

2

If NO, let c 2 = 0, a 2 = a 1

bin

c 1 c 2 , b 2 = m 1 = a 2

2

. If YES, let c 2

a 2 = m 1

bin

c 1 c 2 , b 2 = b 1 = a 2

2

In both cases, for n ≥ n 2 , a 2 = 0. bin

c 1 c 2 ≤ xn < b 2 = a 2 +

2

and b 2 − a 2 =

2

chap7c.pdf page 1/

By recursion (on k), if n k

n k− 1 , c 1 ,... , c k , a k

bin

c 1

... c k , b k = a k

k

have

been defined so that for n ≥ n k

a k

bin

c 1

... c k ≤ x n < b k = a k

k

divide the interval [a k , b k ) into two halves.

Is there an n k+

n k such that x nk+ ≥ m k = a k

k+

If NO, let c k+ = 0, a k+ = a k

bin

c 1 c 2

... c k+ , b k+ = m k = a k+

k+

. If

YES, let c k+ = 1, a k+ = m k

bin

c 1 c 2

... c k+ , b k+ = b k = a k+

k+

In both cases, n k+

n k , c 1 ,... , c k , c k+ , a k+

bin

c 1

... c k c k+ , b k+ = a k+

k+

have been defined so that for n ≥ n k+

a k+

bin

c 1

... c k+ ≤ x n < b k+ = a k+

k+

Let

L = 0.

bin

c 1... ck...

= lim

k→∞

a k

= lim

k→∞

b k

We have that L = limn→∞ xn since for all k, and n > nk,

0 ≤ L − x n ≤ b k − x n ≤ b k − a k

k

chap7c.pdf page 2/