MT2002 Analysis

 Previous page (Some properties of convergent sequences) Contents Next page (Subsequences)

## Monotonic sequences

We now want to investigate what the Completeness axiom tells us about the convergence of sequences.
Unfortunately, the example of the sequence (1, 0, 1, 0, ... ) shows that bounded sequences do not necessarily have limits.

We need the following.

Definition

A sequence (an) is monotonic increasing if an+1 an for all n N.

Remarks

The sequence is strictly monotonic increasing if we have > in the definition.
Monotonic decreasing sequences are defined similarly.

Then the big result is

Theorem

A bounded monotonic increasing sequence is convergent.

Proof
We will prove that the sequence converges to its least upper bound (whose existence is guaranteed by the Completeness axiom).
So let be the least upper bound of the sequence. Given > 0, we'll show that all the terms of the sequence (except the first few) are in the interval ( - , + ).
Now since + is an upper bound of the sequence, all the terms certainly satisfy an< + .
Also since - is not an upper bound of the sequence, we must have aN > - for some N. But then all the later terms will be > - also and so (for n > this N) we have our condition for convergence.

Remarks

A similar result is true for a bounded monotonic decreasing sequence (which converges to its greatest lower bound).

There are sequences which are convergent without being monotonic.
For example, the sequences (-1 , 1/2 , -1/3 , 1/4 , ... ) and (1/2 , 1/22 , 1/3 , 1/32 , ... ) both converge to 0.

For sequences given by recurrence relations it is sometimes easy to see what their limits are.

Examples

1. Define a sequence by an+1= (2an- 1) with a1= 2.

Then suppose that this has a limit . Then for large n, we have an= (approx) and an+1= (approx) and so we must have = (2 -1) and hence 2 = 2 - 1 and we get = 1.

To show that it does indeed have a limit, we'll prove that it is monotonic decreasing and bounded below.

Since the terms of the sequence are positive, the sequence is clearly bounded below by 0.

To show monotonicity:
an+1- an= (2an- 1) - (2an-1- 1)
and if we now use the fact that x - y = (x - y)/(x + y)
we get an+1- an= 2(an-an-1)/((2an- 1) + (2an-1- 1)) and the result then follows by induction.

2. Define an+1= (an2 + 1)/2 with a1 = 2.

Claim: This is monotonic increasing.
an+1 - an= (an2 - an-12)/2 = (an + an-1)(an - an-1)/2 and so (since an > 0) the result follows by induction.

What is the limit ?
The limit satisfies = (2+ 1)/2 and hence = 1.
But since the sequence starts at 2 and increases, it cannot converge to 1 and hence it has no limit.

One may show that if one takes the starting value a1 = 1/2 then the sequence is bounded above and does converge to 1.

 Previous page (Some properties of convergent sequences) Contents Next page (Subsequences)

JOC September 2002