Sequences and convergence

Sequences

Definition 1(Definition).

A sequence is a function whose domain is $\mathbb{N}$.

Given a function $f:\mathbb{N}\to \mathbb{R}$, $f(n)$ is just the $n$th term of the sequence.

It is evident that sequences must be infinitely long.

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Definition of convergence in real numbers and metric spaces

Definition 2(Definition).

A sequence $(a_n)$ in $\mathbb{R}$ converges to $a\in \mathbb{R}$ if, for every positive number $ϵ$, there exists an $N\in \mathbb{N}$ such that whenever $n\ge N$ it follows that $|a_n-a|<ϵ$.

A sequence $(a_n)$ in a metric space $(X,d)$ converges to $a\in X$ if, for every positive number $ϵ$, there exists an $N\in \mathbb{N}$ such that whenever $n\ge N$ it follows that $d(a_n,a)<ϵ$.

Alternatively, $(a_n)$ converges to $a\in X$ if and only if every neighborhood of $a$ contains $a_n$ for all but finitely many $n$.

All these

• $\lim a_n=a$
• ${\lim }_{n\to \infty }{a}_n=a$
• $a_n\to a$

indicate that $(a_n)$ converges to $a$.

Definition 3(Definition).

A sequence that does not converge is said to diverge.

Template for a proof that $(x_n)\to x$

• Let $ϵ>0$ be arbitrary.
• Demonstrate a choice for $N\in \mathbb{N}$. This is usually done in terms of $ϵ$.
• Show that $N$ works, i.e, for all $n\ge N$, $|x_n-x|<ϵ$.

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Uniqueness of a limit

Theorem 4(Theorem).

The limit of a sequence, when it exists, is unique.

Proof
Let $(a_n)\to a$ and $(a_n)\to b$. FSTOC, assume $a\ne b$. WLOG, $a<b$. Let $ϵ<\frac{b-a}{2}$.
Now, there must exist an $N$ such that for all $n\ge N$, $|a_n-a|<ϵ$. From the triangle inequality, we have

$$\begin{array}{rl}|b-a|& \le |a_n-b|+|a_n-a|\\ 2ϵ& <|a_n-b|+ϵ\\ |a_n-b|& >ϵ,\forall n\ge N\end{array}$$

This contradicts our hypothesis that $(a_n)\to b$. $\Rightarrow \Leftarrow$ ❏

Another way to prove this would be to use the fact that there exist $N_1,N_2\in \mathbb{N}$ such that for any arbitrary $ϵ>0$

$$\begin{array}{rl}|a_n-a|& <\frac{ϵ}{2}\forall n\ge N_1\text{ and}\\ |a_n-b|& <\frac{ϵ}{2}\forall n\ge N_2\end{array}$$

Let $N=\max \{N_1,N_2\}$. From the triangle inequality, we have

$$\begin{array}{rl}|a-b|& \le |a-a_n|+|a_n-b|\\ & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.\end{array}$$

$|a-b|<ϵ$ for any positive $ϵ$. This can only happen if $a=b$.

This proof can be modified to work for a general metric space when we use open balls of diameter $ϵ$ centered at $a$ and $b$ in place of open intervals of size $ϵ$.

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Boundedness

Definition 5(Definition).

A sequence $(x_n)$ in $\mathbb{R}$ is bounded if there exists a number $M>0$ such that $|x_n|\le M$ for all $n\in \mathbb{N}$.

A sequence $(x_n)$ in a metric space $(X,d)$ is bounded if it is contained in a ball.

Theorem 6(Theorem).

Every convergent sequence is bounded.

Proof
Assume $(x_n)\to l$. This means that given a particular value of $ϵ$, we know that there must exist an integer $N\in \mathbb{N}$ such that if $n\ge N$, $x_n\in (l-ϵ,l+ϵ)$. Thus, the sequence is bounded for $n\ge N$ by $|l|+ϵ$. We still need to worry about the terms in the sequence that come before the $N$th term. Because there are only a finite number of these, we let

$$M=\max \{|x_1|,|x_2|,|x_3|,\dots ,|x_{N-1}|,|l|+ϵ\}.$$

It follows that $|x_n|\le M$ for all $n\in \mathbb{N}$. ❏

This proof can be easily rephrased in terms of open balls to prove the theorem for metric spaces.

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Limit theorems

Algebraic limit theorem

Theorem 7(Theorem).

Let ${\lim }_{n\to \infty }{a}_n=a$, and ${\lim }_{n\to \infty }{b}_n=b$. Then,

1. ${\lim }_{n\to \infty }(c{a}_n)=ca$, for all $c\in \mathbb{R}$;
2. ${\lim }_{n\to \infty }(a_n+b_n)=a+b$;
3. ${\lim }_{n\to \infty }(a_n{b}_n)=ab$;
4. ${\lim }_{n\to \infty }\left(\frac{a_n}{b_n}\right)=\frac{a}{b}$, provided $b\ne 0$.

Proof
Algebraic Limit Theorem in Vector Spaces

Order limit theorem

Theorem 8(Theorem).

Let ${\lim }_{n\to \infty }{a}_n=a$, and ${\lim }_{n\to \infty }{b}_n=b$. Then,

1. If $a_n\ge 0$ for all $n\in \mathbb{N}$, then $a\ge 0$.
2. If $a_n\le b_n$ for all $n\in \mathbb{N}$, then $a\le b$.
3. If there exists $c\in \mathbb{R}$ for which $c\le b_n$ for all $n\in \mathbb{N}$, then $c\le b$. Similarly, if $a_n\le c$ for all $n\in \mathbb{N}$, then $a\le c$.

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Monotone convergence theorem

Definition 9(Definition).

A sequence $(a_n)$ is increasing if $a_n\le a_{n+1}$ for all $n\in \mathbb{N}$ and decreasing if $a_n\ge a_{n+1}$ for all $n\in \mathbb{N}$. A sequence is monotone if it is either increasing or decreasing.

Info

Note that this only applies to spaces in which an order is defined. Does not make sense in vanilla metric spaces. We will stick to $\mathbb{R}$.

Theorem 10(Theorem).

If a sequence is monotone and bounded, it converges.

More formally, if $(p_n)$ is a monotone bounded sequence with $p=\sup \{p_1,p_2,\dots \}$ if $(p_n)$ is increasing and $p=\inf \{p_1,p_2,\dots \}$ if $(p_n)$ is decreasing, then $(p_n)\to p$.

Proof
Assume $(p_n)$ is increasing. We have to prove that $(p_n)\to p$. Let $S=\{p_1,p_2,\dots \}$. Consider the number $p-ϵ$ for arbitrary $ϵ>0$. Since $p$ is $\sup S$, there must exist $p_N\in S$ such that

$$\begin{array}{rl}& p-ϵ<p_N\\ ⟹& |p-p_N|<ϵ.\end{array}$$

Since $(p_n)$ is a monotone increasing sequence, the above inequality is also true for all $n\ge N$. A similar argument works for decreasing $(p_n)$. ❏

Theorem 11(Theorem).

If a sequence is monotone and convergent, it is bounded.

Proof
Assume $(p_n)\to p$ is increasing (the decreasing case is handled similarly). Clearly, $p_1$ is a lower bound for $S=\{p_1,p_2,\dots \}$.
Next, FTSOC, assume $p_k>p$ for some $k$. It follows that $p<p_k\le p_n$ for all $n\ge k$. Now, if we set $ϵ=p_k-p$, we get $p_n-p\ge ϵ$ for all $n\ge k$, i.e, the sequence does not converge. $\Rightarrow \Leftarrow$
Thus, $p_k\le p$ for all $k$, i.e, $p$ is an upper bound. ❏

So, just to recap:

• convergent $⟹$ bounded.
• bounded $/⟹$ convergent. Ex 1, -1, 1, -1, …
• monotone $/⟹$ convergent. Ex 1, 2, 3, 4, …
• If a sequence is monotone, bounded $⟺$ convergent.

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Squeeze theorem

Theorem 12(Theorem).

If $(x_n)\to a$, $(z_n)\to a$ and $x_n<y_n<z_n$ for all sufficiently large $n$, then $(y_n)\to a$.

Proof
Let $ϵ>0$ be arbitrary. Then, there must exist an $N_1$ such that for all $n\ge N_1$, $|x_n-a|<ϵ$, which implies $x_n>a-ϵ$. Similarly, there must exist an $N_2$ such that for all $n\ge N_2$, $|z_n-a|<ϵ$, which implies $z_n<a+ϵ$. Thus, $a-ϵ<y_n<a+ϵ$, i.e, $|y_n-a|<ϵ$. ❏

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Limits of some special sequences

Theorem 13(Theorem).

1. If $p>0$, then ${\lim }_{n\to \infty }\frac{1}{n^p}=0$.
2. If $p>0$, then ${\lim }_{n\to \infty }\sqrt[n]{p}=1$.
3. ${\lim }_{n\to \infty }\sqrt[n]{n}=1$.
4. If $p>0$ and $\alpha$ is real, then ${\lim }_{n\to \infty }\frac{n^{\alpha }}{(1+p{)}^n}=0$.
5. If $|x|<1$, then ${\lim }_{n\to \infty }{x}^n=0$.

Check Rudin, 3.20 for proof.

Note

In (b), Rudin says

$$p=(1+x_n{)}^n\ge 1+n{x}_n.$$

In (c), Rudin says

$$n=(1+x_n{)}^n\ge \frac{n(n-1)}{2}{x}_n^2.$$

Notice that the upper bound used in (b) is also valid as an upper bound in (c):

$$n=(1+x_n{)}^n\ge 1+n{x}_n.$$

But, this is a useless bound, since it gives

$$0\le x_n\le \frac{n-1}{n},$$

where the upper bound does not converge to $0$. Thus, we need to use a stronger upper bound, like the one Rudin uses.