Question

In the sequence anif an+1 ≥ an for all values of n, the sequence is called as

Oscillatory

Infinite

Monotonically increasing

Monotonically decreasing ​

Answer 1

Answer:

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 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.