Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
The Monotone Sequence Theorem, which guarantees the convergence of increasing and decreasing sequences that are bounded above or below, respectively. The theorem is proven using the concept of sup and the Least-Upper Bound Property. The document also covers the concept of lim sup and its relationship to the Monotone Sequence Theorem.
Typology: Slides
1 / 22
This page cannot be seen from the preview
Don't miss anything!
Video: Monotone Sequence Theorem
Notice how annoying it is to show that a sequence explicitly converges, and it would be nice if we had some easy general theorems that guar- antee that a sequence converges.
Definition: (sn) is increasing if sn+1 > sn for each n (sn) is decreasing if sn+1 < sn for each n If either of the above holds, we say that (sn) is monotonic.
Date: Monday, April 20, 2020.
1
Examples: sn =
n is increasing, sn = (^1) n is decreasing, sn = (−1)n is neither increasing nor decreasing.
The following theorem gives a very elegant criterion for a sequence to converge, and explains why monotonicity is so important.
Monotone Sequence Theorem:
(sn) is increasing and bounded above, then (sn) converges.
Note: The same proof works if (sn) is nondecreasing (sn+1 ≥ sn)
Intuitively: If (sn) is increasing and has a ceiling, then there’s no way it cannot converge. In fact, try drawing a counterexample, and you’ll see that it doesn’t work!
We need to find N such that if n > N , then |sn − s| < .
Consider s − < s. By definition of a sup, this means that there is sN ∈ S such that sN > s −
But then, for that N , if n > N , since sN is increasing, we have
sn − s > sN − s > −
On the other hand, since s = sup(S) by definition of sup, we have sn ≤ s for all s and so
sn − s ≤ s − s = 0 <
Therefore we get
− < sn − s < ⇒ |sn − s| <
And so (sn) converges to s
Of course, by considering −sn we get the following corollary:
Corollary:
(sn) is decreasing and bounded below, then (sn) converges.
Why? In that case (−sn) is increasing and bounded above, so con- verges to s, and therefore (sn) converges to −s (or repeat the above
Notice first of all that there is N such that sN > M , because otherwise sN ≤ M for all N and so M would be an upper bound for (sn).
With that N , if n > N , then since (sn) is increasing, we get sn > sN = M , so sn > M and hence sn goes to ∞ X
Finally, notice that the proof of the Monotone Sequence Theorem uses the Least-Upper Bound Property (because we defined sup), but in fact something even more awesome is true:
Cool Fact: The Least Upper Bound Property is equivalent to the Monotone Sequence Theorem! (WOW)
Video: Decimal Expansions
There is a more natural (but less elegant) construction of R than us- ing cuts that you’re probably more acquainted with, namely decimal expansions.
Motivation: What does is mean for π = 3. 1415 · · ·?
Notice:
π =3 +
=k +
d 1 10
d 2 102
d 3 103
Now consider the following sequence (sn)
s 0 =3 = k
s 1 =3.1 = k +
d 1 10 s 2 =3.14 = k +
d 1 10
d 2 102 s 3 =3.141 = k +
d 1 10
d 2 102
d 3 103 sn =3. 1415 · · · dn = k +
d 1 10
d 2 102
dn 10 n
Notice that (sn) is bounded above by 4 = k + 1 and moreover, (sn) is increasing (since we’re only adding positive terms), therefore by the monotone sequence theorem, (sn) is converging to s, and this limit is what we call
9 10
So actually we have 0. 999999 · · · = 1. 00000 · · · , so both of those deci- mal expansions actually represent the same real number! So the above construction is bad in the sense that different decimal expansions might give you the same number. This is different for cuts; different cuts ac- tually give different real numbers!
There is an easy way to get around that, actually: In the above con- struction, simply throw away decimal expansions that end with an infinite string of 9′s. That is, in the above definition, consider just the decimal expansions that don’t end with 9′s.
Video: What is lim sup?
Finally, let me discuss the second most important concept in analysis (after sup of course): The lim sup. Because so far we talked about con- vergent sequences. But in reality, a lot of sequences don’t converge! How do we deal with them?
Note: For the following, assume that (sn) is bounded (but see below).
Consider the following example:
Even though (sn) doesn’t converge, we would like to say that the largest possible limit (= limsup) of (sn) is 1 and the smallest possible limit (= liminf) of (sn) is −1.
Notice that the lim sup is NOT the same as the sup. In this example, the sup is 4 but the limsup is 1.
The idea is as follows: The limsup of sn is essentially the sup of sn, but for large values of n.
Although things generally don’t always stabilize, what is true is that:
Fact: (vN ) is a decreasing sequence
Why? For example, notice that for v 0 = sup {sn | n > 0 } you have lots of values of sn to compare, and here in fact the sup is 4. But for vN = sup {sn | n > N } you have much fewer values to compare, so the sup cannot be as big as the original one!
Interpretation: All this means is that the lim sup is the sup of sn but for large values of n, so it’s really essentially the largest possible limit of sn
Example:
Find lim supn→∞ sn where sn = (−1)n
Notice that for every N (not necessarily large),
vN = sup {sn | n > N } = 1
And therefore
lim sup n→∞
sn = lim N →∞
sup {sn | n > N } = lim N →∞
And why is lim sup SO important? Because even though limn→∞ sn doesn’t always exist, we have:
Upshot: lim supn→∞ sn ALWAYS exists!
And it’s GOOD for things to exist! For example, notice how useful it is for sup(S) to always exist (we used that a LOT), and same goes for lim sup.
Note: So far we assumed that (sn) is bounded, but even if it’s not we say that:
Definition: If (sn) is not bounded above, then we define
lim sup n→∞
sn = ∞
And this time define (uN ) by:
uN = inf {sn | n > N }
(So this time you look at the smallest value of sn after N ) Let’s plot a couple of values of uN
u 0 = inf {sn | n > 0 } = − 4 u 1 = inf {sn | n > 1 } = − 3 u 2 = − 2 u 3 = − 1 u 4 = − 1 u 5 = − 1
And just as before, we have:
Fact: (uN ) is an increasing sequence
Why? Again, it’s because we have fewer and fewer values to compare, which causes to increase the inf.
