Honors Calculus Notes for [1 Sequences]

Definition

Def. 1.1 Limit of a sequence

If the following holds:

\(\forall \epsilon>0,\exists N>0\) such that if \(n>N,\) then \(|a_n-L|<\epsilon\)

Def. 1.2 Subsequences

Let \(\lbrace a_n\rbrace_{n=1}^\infty\) be a sequence of real numbers. A subsequence of \(\lbrace a_n\rbrace_{n=1}^\infty\) is in the form \(\lbrace a_{n_k}\rbrace_{k=1}^\infty\), where \(n_k\in \mathbb N\) for all \(k\in\mathbb N\), and

\[\forall i\in \mathbb N, n_i<n_{i+1} \]

Def. 1.3 Monotone sequences

A sequence \(\lbrace a_n\rbrace\) is said to be:

• monotone increasing if \(a_n\leq a_{n+1}\) for any \(n\in\mathbb N\)

• strictly increasing if \(a_n< a_{n+1}\) for any \(n\in\mathbb N\)

• monotone decreasing if \(a_n\geq a_{n+1}\) for any \(n\in\mathbb N\)

• strictly decreasing if \(a_n> a_{n+1}\) for any \(n\in\mathbb N\)

Def. 1.4 Bounded sequences

A sequence \(\lbrace x_n\rbrace\) is said to be bounded from above if there exists a constant \(\beta\in\mathbb R\) such that \(x_n\leq \beta\) for any \(n\in \mathbb N\). It is said to be bounded from below if there exists a constant \(\alpha\in\mathbb R\) such that \(x_n\geq \alpha\) for any \(n\in \mathbb N\).

If the sequence is both bounded from above and from below, we may simply say the sequence is bounded

Def. 1.5 Euler's number

\[e:=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n \]

Def. 1.6 Cauchy sequences

\(\lbrace a_n\rbrace\) is said to be a a Cauchy sequence if the following holds

\(\forall \epsilon>0, \exists N>0\) such that if \(m,n>N\), then \(|a_m-a_n|<\epsilon\).

Def. 1.7 Infinity

Given a sequence of real numbers \(\lbrace a_n\rbrace\), if \(\forall K>0, \exists N>0\) such that \(\forall n>N, a_n>K\), we say \(\lim_{n\rightarrow\infty} a_n=+\infty\).

Likewise, if \(\forall K>0, \exists N>0\) such that \(\forall n>N, a_n<-K\), we say \(\lim_{n\rightarrow\infty} a_n=-\infty\).

Proposition & Theorem

Pro. 1.1 Arithmetic Rules

We have:

1. \(\lim_{n\rightarrow\infty}c=c\) where \(c\) is a constant.

2. \(\lim_{n\rightarrow\infty}\frac{1}{n^p}=0\) where \(p>0\) is a fixed positive number.

3. If \(\lim_{n\rightarrow\infty} a_n=L\) and \(\lim_{n\rightarrow\infty} b_n=M\) where \(L\) and \(M\) are finite numbers, then

   (a) \(\lim_{n\rightarrow\infty} a_n\pm b_n=L\pm M\)

   (b) \(\lim_{n\rightarrow\infty}a_nb_n=LM\)

   (c) Suppose that \(M\neq 0\), then \(\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac{L}{M}\)

Proof 2

\[n>\frac{1}{\epsilon^{\frac{1}{p}}} \]

Proof 3.(a)

\[|(a_n+b_n)-(L+M)|=|(a_n-L)+(b_n-M)|\leq |a_n-M|+|b_n-M|<\frac{\epsilon}{2}+\frac{\epsilon}{2} \]

Proof 3.(b)

\[|a_nb_n-LM|=|(a_n-L)b_n+L(b_n-M)|\leq |a_n-L||b_n|+|L||b_n-M| \]

Pro. 1.2 Squeeze Theorem

If there exists \(N>0\) such that \(a_n\leq b_n\leq c_n\) for any \(n\geq N\), and

\[\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}c_n=L \]

where \(L\in \mathbb R\) (i.e. finite), then \(\lbrace b_n\rbrace\) also has limit and \(\lim_{n\rightarrow\infty}b_n=L\).

Proof Pro. 1.2 Squeeze Theorem

\(\forall \epsilon>0\), \(\exists N_1>0\) such that \(\forall n>N_1, |a_n-L|<\epsilon\). In particular we have

\[-\epsilon<a_n-L<\epsilon \]

Similarly, \(\exists N_2>0\) such that if \(n>N_2\), we have

\[-\epsilon<c_n-L<\epsilon \]

\(\therefore \exists N^{\prime}=\max\lbrace N, N_1, N_2\rbrace\), such that \(\forall n>N^{\prime}\), we have

\[-\epsilon<a_n-L<b_n-L<c_n-L<\epsilon \]

This implies \(|b_n-L|<\epsilon\).

\(\therefore \lim_{\rightarrow\infty} b_n=L\)

Pro. 1.3

Given and sequence \(\lbrace a_n\rbrace\) of real numbers, we have

\(\lim_{n\rightarrow\infty}a_n=0\Leftrightarrow\lim_{n\rightarrow\infty}|a_n|=0\)

Pro. 1.4 Comparison rules

Suppose \(\lim_{n\rightarrow\infty} a_n\) and \(\lim_{n\rightarrow\infty} b_n\) exist, then the following implications hold:

(a) If \(\lim_{n\rightarrow\infty}a_n<\lim_{n\rightarrow\infty} b_n\), then there exists \(N>0\) such that \(\forall n>N, a_n<b_n\).

(b) If there exist infinitely many \(n\)'s such that \(a_n\geq b_n\) , then \(\lim_{n\rightarrow\infty}a_n\geq \lim_{n\rightarrow\infty}b_n\).

Proof Pro. 1.4

(b) is the contrapositive of (a).It suffices to prove (a).

Denote \(\lim_{n\rightarrow\infty}a_n=L\) and \(\lim_{n\rightarrow\infty}b_n=M\), and given that \(L<M\).

Pick \(\epsilon=\frac{M-L}{2}>0\), \(\exists N_1>0\) such that if \(n>N_1\), then

\[|a_n-L|<\epsilon=\frac{M-L}{2}\Rightarrow L-\frac{M-L}{2}<a_n<L+\frac{M-L}{2} \]

Similarly, \(\exists N_2>0\) such that if \(n>N_2\), then

\[|b_n-M|<\epsilon=\frac{M-L}{2}\Rightarrow M-\frac{M-L}{2}<b_n<M+\frac{M-L}{2} \]

$\therefore $ when \(n>\max\lbrace N_1, N_2\rbrace\), we have

\[a_n<\frac{M+L}{2}<b_n \]

Pro. 1.5

If \(\lim_{n\rightarrow\infty} a_n=L\), where \(L\in \mathbb R\), then any of its subsequence \(\lbrace a_{n_k}\rbrace_{k=1}^\infty\) would also have

\[\lim_{k\rightarrow\infty}a_{n_k}=L \]

Proof Pro. 1.5

The face is that \(n_k\geq k\) for any \(k\in \mathbb N\).

For any \(\epsilon >0\), \(\exists N>0\) such that \(\forall n>N, |a_n-L|<\epsilon\)

\(\therefore \forall k>N, n_k\geq k>N, |a_{n_k}-L|<\epsilon\)

Corollary 1.6

Given a sequence \(\lbrace a_{n}\rbrace_{n=1}^\infty\) of real numbers, if any of the following holds:

• there exist two subsequences \(\lbrace a_{n_k}\rbrace_{k=1}^\infty\) and \(\lbrace a_{m_j}\rbrace_{j=1}^\infty\) with different limits, or

• there exists a subsequence \(\lbrace a_{n_k}\rbrace_{k=1}^\infty\) whose limit does not exist,

then the limit of \(\lbrace a_{n}\rbrace_{n=1}^\infty\) does not exist.

Pro. 1.7

Any convergent sequence is bounded.

Proof Pro. 1.7

Denote that \(\lim_{n\rightarrow\infty}a_n=L\).

when \(\epsilon=1>0, \exists N>0\) such that \(\forall n>N, |a_n-L|<\epsilon\), that is

\[L-1<a_n<L+1 \]

\(\therefore \forall n\in\mathbb N, \min\lbrace\min_{i=1}^{N} a_i, L-1\rbrace\leq a_n\leq\max\lbrace\max_{i=1}^{N} a_i, L+1\rbrace\)

$\therefore \lbrace a_n\rbrace $ is bounded.

Theo. 1.8

Any monotone and bounded sequence must converge.

Theo. 1.9 Cauchy completeness of \(\mathbb R\)

$\lbrace a_n\rbrace $ is a Cauchy sequence in \(\mathbb R\) if and only if \(\lim_{n\rightarrow\infty} a_n\) exists in \(\mathbb R\).

Lemma 1.10

Every Cauchy sequence is bounded.

Proof Lemma 1.10

Denote \(\lbrace a_n\rbrace\) is a Cauchy sequence.

When \(\epsilon =1>0\), \(\exists N\in\mathbb N\) such that when \(m,n\geq N\), we have \(|a_m-a_n|<1\).

In particular, whenever \(n\geq N\), we have \(|a_N-a_n|<1\Rightarrow -1+a_N<a_n<1+a_N\).

\(\therefore \min\lbrace\min_{i=1}^{N-1}a_i, -1+a_N\rbrace\leq a_n\leq \max\lbrace\max_{i=1}^{N-1}a_i, 1+a_N\rbrace\).

Theo. 1.11 Bolzano-Weiestrass

Any bounded sequence \(\lbrace a_n\rbrace\) must have at least one subsequence \(\lbrace a_{n_k}\rbrace\) that converges.

Lemma 1.12

Any sequence \(\lbrace a_{n}\rbrace\) must have a monotone subsequence \(\lbrace a_{n_k}\rbrace\).

Proof Lemma 1.12

• Define \(a_k\) is a peak if \(\forall i\geq k, a_i\leq a_k\).

Case 1: If \(\lbrace a_n\rbrace\) has infinitely many peaks \((a_{n_1}, a_{n_2}, a_{n_3}, \cdots)\), where \(n_1<n_2<n_3<\cdots\).

Then, \(\lbrace a_{n_k}\rbrace\) is a monotone decreasing subsequence.

Case 2:Otherwise.

There must exist \(N\in \mathbb N\) such that \(a_n\) is not a peak for all \(n\geq N\).

Then we select some indices as follows.

• \(m_1=N\)

• \(\forall i\geq 1\), there exists at least one index \(t>m_i\) such that \(a_t>a_{m_i}\), as \(a_{m_i}\) is not a peak.

This implies \(a_{m_1}<a_{m_2}<a_{m_3}<\cdots\), where \(m_1<m_2<m_3<\cdots\).

Then, \(\lbrace a_{m_j}\rbrace\) is a monotone increasing subsequence.

Proof Theo. 1.11

Any monotone and bounded sequence must converge. [Theo 1.8]

\(\because\lbrace a_n\rbrace\) is a bounded sequence, and \(\lbrace a_{n_k}\rbrace\) or \(\lbrace a_{m_j}\rbrace\) is a monotone subsequence. [Lemma 1.12]

\(\therefore \lbrace a_n\rbrace\) has a bounded and monotone subsequence.

\(\therefore \lbrace a_n\rbrace\) has a subsequence that converges.

Proof Theo. 1.9

Denote \(\lbrace a_n\rbrace\) is a Cauchy sequence.

\(\lbrace a_n\rbrace\) is bounded. [Lemma 1.10]

\(\lbrace a_n\rbrace\) has a subsequence \(\lbrace a_{n_k}\rbrace\) such that \(\lim_{k\rightarrow\infty} a_{n_k}=L\), where \(L\in\mathbb R\). [Theo. 1.11]

\(\forall \epsilon>0\), \(\exists K\in \mathbb N\) such that \(\forall k\geq K, |a_{n_k}-L|<\frac{\epsilon}{2}\). [Def. 1.1]

\(\forall \epsilon>0\), \(\exists N\in\mathbb N\) such that \(\forall n, m\geq N, |a_m-a_n|<\frac{\epsilon}{2}\). [Def. 1.6]

Pick \(k_0\geq \max\lbrace N, K\rbrace\), then \(n_{k_0}\geq k_0\geq N\), we have both

\[|a_{n_{k_0}}-L|<\frac{\epsilon}{2} \]

and

\[|a_n-a_{n_{k_0}}|<\frac{\epsilon}{2}, \forall n\geq N \]

These show for any \(n\geq N\), we have

\[|a_n-L|=|(a_n-a_{n_{k_0}})+(a_{n_{k_0}}-L)|\leq|a_n-a_{n_{k_0}}|+|a_{n_{k_0}}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \]

Pro 1.13

If \(\lbrace a_n\rbrace\) is monotone increasing and unbounded, then \(a_n\rightarrow+\infty\).

Likewise, if \(\lbrace a_n\rbrace\) is monotone decreasing and unbounded, then \(a_n\rightarrow-\infty\).

Proof Pro. 1.13

\(\forall K>0\), there must exists \(a_N\) such that \(a_N>K\) (otherwise: K would be an upper bound). By monotonicity, \(a_n\geq a_N\) for any \(n\geq N\), and therefore it implies \(a_n>K\) for any \(n\geq K\).

Pro 1.14 Arithmetic rules

• If \(a_n\rightarrow\pm \infty\), then \(\frac{1}{a_n}\rightarrow 0\).

• If \(a_n\rightarrow 0\), then \(|\frac{1}{a_n}|\rightarrow +\infty\).

• If $a_n\rightarrow +\infty $ and \(b_n\rightarrow +\infty\) , then \(a_n+b_n\rightarrow +\infty\).

• If $a_n\rightarrow -\infty $ and \(b_n\rightarrow -\infty\) , then \(a_n+b_n\rightarrow -\infty\).

• If $a_n\rightarrow \infty $ and \(b_n\rightarrow L\in\mathbb R\) , then \(a_n+b_n\rightarrow \infty\).

• If $a_n\rightarrow \infty $ and \(b_n\rightarrow L>0\) , then \(a_nb_n\rightarrow \infty\).

• If $a_n\rightarrow \infty $ and \(b_n\rightarrow L<0\) , then \(a_nb_n\rightarrow -\infty\).

Samples

Smp. 1.

Suppose \(\lbrace a_n\rbrace\) is a sequence of positive numbers.

If \(\lim_{n\rightarrow \infty} a_n=L\geq 0\), then

\[\lim_{n\rightarrow \infty} \sqrt a_n=\sqrt L \]

Smp. 2.

Suppose \(\lbrace a_n\rbrace\) is a sequence of real numbers.

If \(\lim_{n\rightarrow \infty} a_n=L\geq 0\), then

\[\lim_{n\rightarrow \infty} |a_n|=|L| \]

\[\lim_{n\rightarrow \infty} a_n^{\frac{1}{3}}=L^{\frac{1}{3}} \]

Smp. 3.

Suppose \(\lbrace a_n\rbrace\) is a sequence of positive numbers.

If \(\lim_{n\rightarrow \infty} a_n=L\), then

\[\lim_{n\rightarrow \infty} a_n^{\frac{p}{q}}=L^{\frac{p}{q}} \]

for any \(p,q\in\mathbb N^+\)

Smp. 4.

\(\lim_{n\rightarrow\infty}\frac{sin n}{n}=0\)

\(\lim_{n\rightarrow\infty}\frac{1}{\sqrt n}=0\)

\(\lim_{n\rightarrow\infty}\sqrt{n+a}-\sqrt n=0\)

\(\lim_{n\rightarrow\infty} a^\frac{1}{n}=1 (a>0)\)

\(\lim_{n\rightarrow\infty} n^\frac{1}{n}=1\)

\(\lim_{n\rightarrow\infty} a^n=0[a\in(-1, 1)]\)

\(\lim_{n\rightarrow\infty} na^n=0[a\in(-1, 1)]\)

\(\lim_{n\rightarrow\infty} \frac{p(n)}{a^n}=0\), \(a>1\) and \(p(x)\) is a polynomial

\(\lim_{n\rightarrow\infty} p(n)^\frac{1}{n}=1\), \(p(x)\) is a polynomial, and \(\forall n\in\mathbb N, p(n)\geq 1\)

Smp. 5.

Suppose \(\lbrace a_n\rbrace\) is a sequence of positive numbers such that

\[\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=L \]

where \(L\in [0,1)\), then

\[\lim_{n\rightarrow\infty} a_n=0 \]