Section 10

This section has two theorems that allow you to know a sequence is converging without knowing the limit prior.

Increasing / decreasing

I will skip super formal definitions, but an increasing sequence is one where the next iteration is always greater or equal to the last. $s_n\le s_{n+1}$. Decreasing is the opposite.

Monotone or monophonic mean consistently increasing or decreasing.

All bounded monotone (increasing) sequences converge.

Basically, if its bounded and the sequence $s_n$ keeps increasing, we can take any $ϵ>0$ and show that for the $\sup s_n$ we can take $\sup s_n-ϵ$ and always define some $N$ where the values are within that rage. Thus the limit is $\sup s_n$.

Think of a limit increasing, it has to be less than the supremum, but it also has to keep going up. For any epsilon there is always going to be an $N$ where any $n>N⟹|s_n-\sup s_n|<ϵ$

Similar logic applies for decreasing and bounded sequences.

10.4:

An expected result is if a sequence is unbounded and increasing it goes to $\infty$. Unbounded and decreasing goes to $-\infty$.

Because of this, the limit of any monotone sequence is meaningful as it either converges, or diverges to an infinity.

limit of a sup and inf

We define the following: $\lim \sup s_n={\lim }_{N\to \infty }\sup \{s_n:n>N\}$ $\lim \inf s_n={\lim }_{N\to \infty }\inf \{s_n:n>N\}$

Here is how to think about it, we are cutting of more and more of the sequence $s_n$ and taking the supremum. Its like asking what the limit is of the list of supremums of each point after all points previous to it has been cut off. The supremum and infimum of any tail we take.

We do not specify $s_n$ to be bounded, but if it is not bounded by above or bellow the supremum or infimum are $\infty$ or $-\infty$ respectfully.

Not in the book, but some examples to understand them better

Suppose $s_n=(0,1,2,1,0,1,2,\mathrm{1....})$

This is just 0,1,2,1 repreating. Note that the $\lim \sup s_n$ is 2 as no matter how many element we cut off the supremum is still 2 Similarly, $\lim \inf$ is 0.

Good example to show differences

Take the sequence $a_n=\frac{1+(-1{)}^n}{n}$ Firstly please note this has different values for odd and even n.

First find $\sup a_n$ and $\inf a_n$.

Then find $\lim \sup n$ and $\lim \inf n$.

todo

This is the limit of the supremum vs. $\sup \{s_n:n>N\}$ being just the supremum. The limit is always smaller than or equal to the supremum. The limit of the supremum is the largest value that infinitely many $s_n$s can get close to.

Build on the last point (i think this leads to squeeze theorem)

1. If $\lim s_n$ is defined, then $\lim \inf s_n=\lim s_n=\lim \sup s_n$.
2. If we have it so $\lim \inf s_n=\lim \sup s_n$ then the limit is defined and $\lim \inf s_n=\lim s_n=\lim \sup s_n$ Defined means it converges or goes to an infinity. Basicly, to know a limit exists we need to show that the limit of the infimum is the same as the limit of the supremum.

This limit also shows that the $\sup \{s_n,n>N\}$ and $\inf \{s_n,n>N\}$ are close together for some large $N$. Also it should now be clear that the $\{s_n,n>N\}$ portion signifies some tail of the sequence after $N$.

Cauchy sequence

A sequence is called an cauncy sequence if For each $ϵ>0$ there exists $N$ such that $n,m>N⟹|s_n-n_m|<ϵ$

(The sequence is converging after a certain $N$ for every $ϵ$).

Convergent sequences are Cauchy sequences

Lets say we have some convergent sequence $s_n$, and let $ϵ>0$.

Then, as it converges, for some $n,m>N$ $|s_n-s|<\frac{ϵ}{2}$ and $|s_m-s|<\frac{ϵ}{2}$

$|s_n-s_m|=|s_n-s+s-s_m|\le |s_n-s|+|s_m-s|<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$

Cauchy sequences are bounded

Very similar proof to one done before

The point:

A sequence is a convergent sequence $⟺$ it is a Cauchy sequence.

More examples

Prove $s_n=\frac{1}{n}$ converges by proving it is Cauchy Scratch work: We need to show that for every $ϵ>0$ there exists an $N$ such that $n,m>N⟹|\frac{1}{n}-\frac{1}{m}|<ϵ$ Starting with the last formula, we can work backwards. Similar to limit proofs. $|\frac{1}{n}-\frac{1}{m}|<ϵ$ We can rewrite this using the triangle formula $|\frac{1}{n}-\frac{1}{m}|\le |\frac{1}{n}|+|-\frac{1}{m}|=|\frac{1}{n}|+|\frac{1}{m}|=\frac{1}{n}+\frac{1}{m}$ Now, if we show that $\frac{1}{n}+\frac{1}{m}<ϵ⟹|\frac{1}{n}-\frac{1}{m}|<ϵ$.

Then, to show this we can show $\frac{1}{n}+\frac{1}{m}<ϵ$ by showing each element is less than $\frac{ϵ}{2}$. For $\frac{1}{n}<\frac{ϵ}{2}$ we need $n>\frac{2}{ϵ}$ For $\frac{1}{m}<\frac{ϵ}{2}$ we need $m>\frac{2}{ϵ}$ Thus we set $N$ to $\frac{2}{ϵ}$.

Formal Proof:

1. Let $ϵ>0$ and $N=\frac{2}{ϵ}$
2. Then, for any $m,n>N$ we have $m,n>\frac{2}{ϵ}$
3. Notice, this means $\frac{1}{n}<\frac{ϵ}{2}$ and $\frac{1}{m}<\frac{ϵ}{2}$
4. By the above inequality, $\frac{1}{n}+\frac{1}{m}<ϵ$
5. Then, notice $|\frac{1}{n}-\frac{1}{m}|\le \frac{1}{n}+\frac{1}{m}<ϵ$
6. Thus for any $ϵ>0$ there exists an $N=\frac{2}{ϵ}$ such that $n,m>N⟹|\frac{1}{n}-\frac{1}{m}|<ϵ$, proving that this sequence is Cauchy.

Squeeze theorem (in hw not book)

Straight out of calc, let $a_n\le s_n\le b_n$ and both $a_n$ and $b_n$ converge to $L$. Then $s_n$ also converges to $L$.