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 (a_n) is monotonic increasing if a_n+1≥ a_n 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 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 a_n< α + ε.
Also since α - ε is not an upper bound of the sequence, we must have a_N> 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.

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 a_n+1= √(2a_n- 1) with a₁= 2.

   Then suppose that this has a limit α. Then for large n, we have a_n= α (approx) and a_n+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:
   a_n+1- a_n= √(2a_n- 1) - √(2a_n-1- 1)
   and if we now use the fact that √x - √y = (x - y)/(√x + √y)
   we get a_n+1- a_n= 2(a_n-a_n-1)/(√(2a_n- 1) + √(2a_n-1- 1)) and the result then follows by induction.

2. Define a_n+1= (a_n² + 1)/₂ with a₁ = 2.

   Claim: This is monotonic increasing.
   a_n+1 - a_n= (a_n² - a_n-1²)/₂ = (a_n + a_n-1)(a_n - a_n-1)/₂ and so (since a_n > 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.
   

---

Previous page (Some properties of convergent sequences)

Contents

Next page (Subsequences)

---

JOC September 2001