aaaaaa −= − + − ≤−+− |||| aa aaronmiech/131A/Hw5/51_7/51_7.pdfaaaaaa aa aa ++ ++++ ++ ++...
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Page 51, Problem 7 Suppose that the terms {a n } satisfy |a n+1 -a n | < 1/2 n . Prove that {a n } is a Cauchy sequence. Proof: To start 1 1 2 2 1 1 2 1 1 1 2 2 | | | | | | | | n n n n n n n n n n n n a a a a a a a a a a + + + + + + + + − = − + − ≤ − + − ≤ + Next, by induction on k, (1) 1 1 1 2 2 | | .... n n k n k n a a + + − − ≤ + + , for 1 1 1 1 1 1 1 2 2 2 | || | | | | | .... n n n n k n k n k k n n k n k n k n k n k n a a a a a a a a a a + + + + + + + + + + + + − − = − + − ≤ − + − ≤ + + + Letting the series in (1) go to infinity we have (2) 1 2 1 1 1 2 2 1 1 1 1 2 2 2 1 1 1 2 1 12 2 | | .... ... ( ...) ( / ) n n k n n k n n n a a + + − − − ≤ + + + = + + + = = − So, given ε > 0, select n so that 1/2 n-1 < ε, and replace a n+k by a m in (2). Then |a m -a n |< ε for m > n. That is {a n } is a Cauchy sequence.
Transcript of aaaaaa −= − + − ≤−+− |||| aa aaronmiech/131A/Hw5/51_7/51_7.pdfaaaaaa aa aa ++ ++++ ++ ++...
Page 51, Problem 7
Suppose that the terms {an} satisfy |an+1 -an| < 1/2n. Prove that {an} isa Cauchy sequence.
Proof: To start
1 12 2
1 12
1
1 12 2
| | | || | | |
n n n n n n
n n n n
n n
a a a a a aa a a a
+ ++ +
+ ++
+
− = − + −≤ − + −
≤ +
Next, by induction on k,
(1) 1
1 12 2
| | ....n nk n k na a+ + −− ≤ + + ,
for
1 1
1
1
1 1 12 2 2
| | | || | | |
....
n n n n k n k nk k
n n k n k nk
n k n k n
a a a a a aa a a a
+ ++ + + +
+ ++ +
+ + −
− = − + −≤ − + −
≤ + + +
Letting the series in (1) go to infinity we have
(2)
1
2
1
1 12 21 1 112 2 21 1 12 1 1 2 2
| | .... ...
( ...)
( / )
n nk n n k
n
n n
a a+ + −
−
− ≤ + + +
= + + +
= =−
So, given εεεε > 0, select n so that 1/2n-1 < εεεε, and replace an+k by amin (2). Then
|am - an| < εεεε for m > n.
That is {an} is a Cauchy sequence.
