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.