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- We now want to investigate what the Completeness axiom tells us about the convergence of sequences.
- 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**- Define a sequence by
*a*_{n+1}= √(2*a*_{n}- 1) with*a*_{1}= 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*= √(2*α*-1) and hence*α*^{2}= 2*α*- 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}= √(2*a*_{n}- 1) - √(2*a*_{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})/(√(2*a*_{n}- 1) + √(2*a*_{n-1}- 1)) and the result then follows by induction. - Define
*a*_{n+1}= (*a*_{n}^{2}+ 1)/_{2}with*a*_{1}= 2.Claim: This is monotonic increasing.

*a*_{n+1}-*a*_{n}= (*a*_{n}^{2}-*a*_{n-1}^{2})/_{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*= (*α*^{2}+ 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

*a*_{1}=^{1}/_{2}then the sequence is bounded above and*does*converge to 1.

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(Subsequences)

- Define a sequence by

Unfortunately, the example of the sequence (1, 0, 1, 0, ... ) shows that bounded sequences do not necessarily have limits.

We need the following.

**Definition**