Definition 2.1. Sequence.
A sequence in a set \(S\) is a function \(x:\mathbb{Z}_{\geq 1}\rightarrow S\text{.}\)
Section 2.1 Sequence Basics
Subsection Sequences and Limits
“A sequence works in a way a collection never can.”―George Murray
Convention 2.2.
We'll usually write \(x_n\) not \(x(n)\text{,}\) and \((x_n)_{n=1}^{\infty}\subseteq S\) in place of \(x:\mathbb{Z}_{\geq 1}\rightarrow S\text{.}\)
Convention 2.3.
Sometimes we'll want a different index set for \(k\in\mathbb{Z}\) also refer to a function \(x:\mathbb{Z}_{\geq k}=\{n\in\mathbb{Z}\,|\,n\geq k\}\rightarrow S\) as a sequence and we'll write \(x\) as \((x_n)_{n=k}^{\infty}\subseteq S\)
Definition 2.4. Limit of a Sequence.
A sequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{C}\) converges to \(x\in\mathbb{C}\) (written \(x_n\rightarrow x\) or \(x=\lim_{n\rightarrow\infty}x_n\)) if for all \(\varepsilon>0\) there exists \(N\in\mathbb{Z}_{\geq 1}\) such that for all \(n\in\mathbb{Z}\) with \(n\geq N\text{,}\) \(|x_n-x|<\varepsilon\text{.}\) (if your sequence starts somewhere else, could have \(N\in\mathbb{Z}_{\geq k}\))
No matter how small a number \(ε\) you pick, so long as it is positive, if you go far enough out in the sequence, all of the terms from that point on will be within a distance of \(ε\) of the limiting value \(L\text{.}\)
Definition 2.5. Convergent, Divergent Sequences.
We say \((x_n)_{n=1}^{\infty}\subseteq S\) converges if there exists some \(x\in\mathbb{C}\) such that \(x_n\rightarrow x\) and we say \((x_n)_{n=1}^{\infty}\subseteq S\) diverges otherwise.
Theorem 2.6. Uniqueness of Sequence Limits.
The limit of a sequence is unique.
Proof.
Suppose \(x_n\rightarrow x\) and \(x_n\rightarrow y\text{.}\) We'll show for all \(\varepsilon>0, |x-y|<\varepsilon\text{.}\) This will show \(|x-y|=0, x=y\text{.}\) To this end, fix \(\varepsilon>0\text{.}\) Since \(x_n\rightarrow x\text{,}\) exists int \(N_1\geq 1\) such that for all \(n\in \mathbb{Z}, n\geq N_1,\) have \(|x_n-x|<\frac{\varepsilon}{2}\text{.}\) Since \(x_n\rightarrow y\text{,}\) exists int \(N_2\geq 1\) such that for all \(n\in \mathbb{Z}, n\geq N_2,\) have \(|x_n-y|<\frac{\varepsilon}{2}\text{.}\) Now let \(N=\max\{N_1,N_2\}\text{.}\) Then since \(n\geq N_1,|x_n-x|<\frac{\varepsilon}{2}\) and since \(n\geq N_2,|x_n-y|<\frac{\varepsilon}{2}\) have \(|x-y|=|(x-x_n)+(x_n-y)|\leq |x-x_n|+|x_n-y|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)
Exploration 2.7.
Compute \(\lim_{n\rightarrow\infty}\frac{3n+5}{4n+7}\text{.}\)
Solution.
Fix \(\varepsilon>0\text{,}\) want to know for all large enough \(n\text{,}\) \(|\frac{3n+5}{4n+7}-\frac{3}{4}|<\varepsilon\text{.}\) Start computing difference, see how big \(n\) needs to be \(|\frac{3n+5}{4n+7}-\frac{3}{4}|=|\frac{4(3n+5)-3(4n+7)}{(4n+7)4}|=\frac{1}{4(4n+7)}\text{.}\) We want \(\frac{1}{4(4n+7)}<\varepsilon\text{.}\) Hold if \(\frac{1}{4n+7}<4\varepsilon\implies 4n+7>\frac{1}{4\varepsilon}\implies n>\frac{1}{4}(\frac{1}{4\varepsilon}-7)\) so as long as \(n>\frac{1}{16\varepsilon}\) have what we want.
Let \(\varepsilon>0\text{.}\) Let \(N\in\mathbb{Z}\) be real int with \(N>\frac{1}{16\varepsilon}\) (exists by Archimedian property of real numbers). Now if \(n\in\mathbb{Z}, n\geq N\) follow equalities and inequalities to \(|\frac{3n+5}{4n+7}-\frac{3}{4}|\leq \dots <\varepsilon\)
Subsection Bounded Sequences
“I am not bound to win, but I am bound to be true.”―Abraham Lincoln
Definition 2.8. Bounded Sequence.
A sequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{C}\) is bounded if exists \(C>0\) such that \(\forall\,n\geq 1, |x_n|\leq C\text{.}\)
Lemma 2.9.
If \(a,b\in\mathbb{C}\) then \(|a-b|\geq ||a|-|b||\)
Proof.
By the triangle inequality,
\begin{equation*}
|a|=|(a-b)+b|\leq |a-b|+|b|
\end{equation*}
so \(|a|-|b|\leq |a-b|\text{.}\) Also \(|b|-|a|\leq |b-a|=|a-b|\) so \(||a|-|b||\leq |a-b|\text{.}\)
Theorem 2.10.
If \((x_n)_{n=1}^{\infty},(y_n)_{n=1}^{\infty}\subseteq \mathbb{C}\text{,}\) \(x_n\rightarrow x\in\mathbb{C}, y_n\rightarrow y\in \mathbb{C}\) then
- \(\displaystyle \lim_{n\rightarrow\infty}(x_n\pm y_n)=x\pm y\)
- \(\displaystyle \lim_{n\rightarrow\infty}(x_ny_n)=x y\)
- If \(y\neq 0,\) exists \(k\in\mathbb{Z}_{\geq 1}\) such that for all \(n\geq k\text{,}\) \(y_n\neq 0\) and sequence \((\frac{x_n}{y_n})_{n=k}^{\infty}\) converges to \(\frac{x}{y}\)
- \(\displaystyle \lim_{n\rightarrow\infty}|x_n|=|x|\)
- \(\displaystyle \lim_{n\rightarrow\infty}\overline{x_n}=\overline{x}\)
- \(\displaystyle \lim_{n\rightarrow\infty}\text{Re}(x_n)=\text{Re}(x)\)
- \(\displaystyle \lim_{n\rightarrow\infty}\text{Im}(x_n)=\text{Im}(x)\)
Theorem 2.11. The Squeeze Theorem.
Suppose \{a_n\}^\infty_{n=1}, {bn}∞ n=1, and {cn}∞ n=1 are three sequences such that \{a_n\}^\infty_{n=1} and {cn}∞ n=1 both converge to L, and an ≤ bn ≤ cn for all n. Then {bn}∞ n=1 also converges to L.
Proof.
Assume \{a_n\}^\infty_{n=1} and {cn}∞ n=1 both converge to L and that an ≤bn ≤ cn for all n ∈ N. We need to prove {bn}∞ n=1 converges to L. Pick ε > 0. Since \{a_n\}^\infty_{n=1} converges to L there is a number N1 such that if n ∈ N and n > N1 then |an−L| < ε and hence L−ε < an < L+ε. Likewise, since {cn}∞ n=1 converges to L there is a number N2 such that if n ∈ N and n > N2 then L − ε < cn < L + ε. Let N = max{N1, N2}. If n ∈ N and n > N, then n > N1 and hence L − ε < an, and n > N2 and hence cn < L + ε, and also an ≤ bn ≤ cn. Combining these facts gives that for n ∈ N such that n > N, we have L − ε < bn < L + ε and hence |bn − L| < ε. This proves {bn}∞ n=1 converges to L.
Theorem 2.12.
Let {an} and {bn} be sequences such that {an} is bounded and limn→∞ bn = 0. Then limn→∞ anbn = 0.
Proof.
There exists a positive number M such that an ≤ M for every positive integer n. Let > 0. There exists a positive integer N such that if n N, then |b_n|<\e/M If n N, then |a_nb_n|<M\e/M=\e.
Subsection Monotonicity
“I'm a stereo and she's just so monotone.”―Nicki Minaj
Definition 2.13. Monotone Sequence.
Suppose \{a_n\}^\infty_{n=1} is any sequence.
- We say \{a_n\}^\infty_{n=1} is increasing if for all n ∈ N, an ≤ an+1. If this inequality is strict, we say \{a_n\}^\infty_{n=1} is strictly increasing.
- We say \{a_n\}^\infty_{n=1} is decreasing if for all n ∈ N, an ≥ an+1. If this inequality is strict, we say \{a_n\}^\infty_{n=1} is strictly decreasing.
- We say \{a_n\}^\infty_{n=1} is monotone if it is either decreasing or increasing. If this inequality is strict, we say \{a_n\}^\infty_{n=1} is strictly monotone.
Warning 2.14.
Be sure to interpret “monotone” correctly. It means (∀n ∈ N, an ≤ an+1) or (∀n ∈ N, an ≥ an+1); it does not mean ∀n ∈ N, (an ≤ an+1) or (an ≥ an+1). Do you see the difference?
Theorem 2.15. Monotone Convergence.
Every bounded monotone sequence converges.
Proof.
It's enough to show every bounded increasing sequence converges. Indeed, if \((x_n)_{n=1}^{\infty}\) is bounded, decreasing, then \((-x_n)_{n=1}^{\infty}\) is bounded and increasing. Also, if \((-x_n)_{n=1}^{\infty}\) then so does \((x_n)_{n=1}^{\infty}\text{.}\)
Now suppose \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\) is increasing and bounded. Since \(\{x_n\,|\,n\geq 1\}\subseteq \mathbb{R}\) is bounded, it has a supremum. Set \(x=\sup\{x_n\,|\,n\geq 1\}\in \mathbb{R}\text{.}\) Show \(x\) is limit. (Remark: we'll often write \(\sup\{x_n\,|\,n\geq 1\}=\sum_{n\geq 1}x_n\)) We'll show \(x_n\rightarrow x\text{.}\) Fix \(\varepsilon>0\text{.}\) Since \(x-\varepsilon<\sup_{n\geq 1}x_n\text{,}\) there is an integer \(N\geq 1\) such that \(x-\varepsilon<x_N\text{.}\) Now for all \(n\geq N\text{,}\) \(x-\varepsilon<x_N\leq x_n\leq x\) so \(|x-x_n|<\varepsilon\,\forall\,n\geq N\) so \(\lim_{n\rightarrow\infty}x_n=x\text{.}\)
Example 2.16.
Define \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\) by \(x_0=0, x_{n+1}=\sqrt{2+x_n}\) for \(n\geq 1\) (so \(x_2=\sqrt{2}, x_3=\sqrt{2+\sqrt{2}}, x_4=\sqrt{2+\sqrt{2+\sqrt{2}}}\text{.}\) We'll find \(\lim_{n\rightarrow \infty}x_n\text{.}\)
Solution.
\((x_n)_{n=1}^{\infty}\) is increasing (strictly, but this not important. First note \(x_1\leq x_2\text{.}\) Further if \(n\geq 1\) and \(x_n\leq x_{n+1}\text{,}\) then
\begin{equation*}
x_{n+1}=\sqrt{2+x_n}\leq \sqrt{2+x_{n+1}}=x_{n+2}.
\end{equation*}
If \(n\geq 1\) and \(x_n\leq 2\) then \(x_n+2\leq 4\) and \(x_{n+1}=\sqrt{x_n+2}\leq 2\) so bounded above by induction. Now we have that \((x_n)_{n=1}^{\infty}\) converges to something. Let \(x=\lim_{n\rightarrow\infty}x_n\text{.}\) Since \(x_{n+1}=\sqrt{x_n+2}\text{,}\) we have \(x=\lim_{n\rightarrow\infty}=\lim_{m\rightarrow\infty}\sqrt{x_n+2}=\sqrt{\lim_{n\rightarrow\infty}x_n+2}=\sqrt{x+2}\text{.}\) Now \(x^2=x+2\) and \(x\geq 0\) so \(x=2\) and \(\lim_{n\rightarrow\infty}x_n=2\text{.}\)
Proposition 2.17. Convergence to Real Numbers.
For any real number r, there exists a strictly increasing sequence of rational numbers that converges to r.
Proof.
We will construct a sequence of rational numbers {qn}∞ n=1 such that r− 1 n < qn < r for every n that is strictly increasing, and then show that this sequence converges to r. By Density of Rational Numbers, there exists a rational number q1 such that r − 1 < q < r. Given q1,..., qn, we define qn+1 recursively to be a rational number such that max{r − 1 n+1, qn} < qn+1 < r again using Density of Rational Numbers. To see that this rule makes sense, we observe that if we have constructed q1,..., qn by this rule, then qn < r, so max{r − 1 n+1, qn} < r, and hence Density of Rational number applies, so we can construct qn+1 (and hence we can construct qn for any n by this rule). Since qn ≤ max{r − 1 n + 1, qn} < qn+1 for every n, the sequence we obtain is strictly increasing. Since r − 1 n < qn < r for every n and {r − 1 n }∞ n=1 converges to r, by the Squeeze Theorem, the sequence {qn}∞ n=1 converges to r.
