MT2002 Analysis

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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.


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


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

Then the big result is


A bounded monotonic increasing sequence is convergent.

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


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.


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

    Then suppose that this has a limit α. Then for large n, we have an= α (approx) and an+1= α (approx) and so we must have α = √(2α -1) and hence α2 = 2α - 1 and we get α = 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 α satisfies α = (α2+ 1)/2 and hence α = 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.

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JOC September 2001