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+1gte an for all n belongs N.


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 alpha be the least upper bound of the sequence. Given epsilon > 0, we'll show that all the terms of the sequence (except the first few) are in the interval (alpha - epsilon, alpha + epsilon).
Now since alpha + epsilon is an upper bound of the sequence, all the terms certainly satisfy an< alpha + epsilon.
Also since alpha - epsilon is not an upper bound of the sequence, we must have aN > alpha - epsilon for some N. But then all the later terms will be > alpha - epsilon 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= sqrt(2an- 1) with a1= 2.

    Then suppose that this has a limit alpha. Then for large n, we have an= alpha (approx) and an+1= alpha (approx) and so we must have alpha = sqrt(2alpha -1) and hence alpha2 = 2alpha - 1 and we get alpha = 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= sqrt(2an- 1) - sqrt(2an-1- 1)
    and if we now use the fact that sqrtx - sqrty = (x - y)/(sqrtx + sqrty)
    we get an+1- an= 2(an-an-1)/(sqrt(2an- 1) + sqrt(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 alpha satisfies alpha = (alpha2+ 1)/2 and hence alpha = 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 2002