Lecture 12 - Monotonic Sequences

See last time for our review of the ALT. See the link for more information.

As a result we get some basic facts:

Facts

  1. If (an)=(c) for some cR then liman=c.
  2. If (an)=(a1,a2,...) and suppose liman=aR. Then any NN has it such that for all nN then (an)a still. See 22 Sequences Practice#2.2.7.

This allows us to prove the Limits and Order (Order Limit Theorem):

Order Limit Theorem

Suppose liman=a and limbn=b. Then:
i. If an0 for all nN, then a0.
ii. If anbn for all nN then ab.
iii. If cR where cbn for all nN, then cb. Similarly, if anc for all nN then ac.

Proof

  1. Suppose an0 for all nN. Assume that a<0 for the contrary. Since ana then for any nN. Then:
    |ana|<ε=aa<ana<a
    But then looking at the right inequality:
    ana<aan<0
    contradicts that an0. As a result, a0 so then a0.

  2. Notice that bnan0 for all nN, so then use (1) along with the Limit Laws (Algebraic Limit Theorem) to get that ba0ba.

  3. Create the sequence (cn)=c for all nN. Then apply (ii) for both cases, giving that since (cn)c via our Limit Laws (Algebraic Limit Theorem), then cb and similarly anc.

Note

We can make the theorem stronger by only imposing that eventually that these sequences have an0 after some N point, for all nN. This would be proved in a similar way.
If a property has this idea, then we say that the sequence eventually has this property.

We then covered Monotonic Sequences:

Monotonic Sequences

A Sequence (an) is

  • increasing if nN we have anan+1
  • decreasing if nN we have anan+1
  • monotone if it is either increasing or decreasing.

See also Monotone Convergence Theorem.