06 Accumulation Points, Sequences in Reals and Complex, and Series

Lecture from: 06.03.2024 | Video: Video ETHZ

This lecture continues our exploration of sequences and introduces the concept of accumulation points, extends our understanding to sequences in higher dimensions and complex numbers, and provides a foundational introduction to series.

Accumulation Points (Häufungspunkte)

Clicker Question and Motivation

Consider the sequence $(a_n{)}_{n\ge 1}$ that is an enumeration of the rational numbers in the interval $[0,1]\cap \mathbb{Q}$.

• What is the limit inferior of $(a_n)$?
• What is the limit superior of $(a_n)$?
• Is $\frac{1}{2}$ a limit of a subsequence of $(a_n)$?

Answers and Explanation:

• Limit Inferior: ${\mathrm{liminf}}_{n\to \infty }{a}_n=0$. Since the sequence contains rational numbers arbitrarily close to 0, the infimum of the tail of the sequence will approach 0.
• Limit Superior: ${\mathrm{limsup}}_{n\to \infty }{a}_n=1$. Similarly, the sequence contains rational numbers arbitrarily close to 1, so the supremum of the tail approaches 1.
• Is $\frac{1}{2}$ a limit of a subsequence? Yes! Since the rational numbers are dense in $\mathbb{R}$, for any $ϵ>0$, the interval $(\frac{1}{2}-ϵ,\frac{1}{2}+ϵ)$ contains infinitely many rational numbers (and thus infinitely many terms of the sequence $(a_n)$).

Demonstration for $\frac{1}{2}$:

For any $n\ge 1$, consider the interval $[\frac{1}{2},\frac{1}{2}+\frac{1}{n}]$. This interval contains infinitely many rational numbers because $\mathbb{Q}$ is dense in $\mathbb{R}$. Therefore, we can always find terms of the sequence $(a_n)$ within this interval. This observation motivates the concept of accumulation points.

Definition of Accumulation Point

Let $(a_n{)}_{n\ge 1}$ be a sequence of real numbers. A real number $c\in \mathbb{R}$ is called an accumulation point (or cluster point, limit point, Häufungspunkt) of $(a_n{)}_{n\ge 1}$ if there exists a subsequence $(a_{n_k}{)}_{k\ge 1}$ that converges to $c$.

Analogously, we say that $\infty$ (or $-\infty$) is an accumulation point of $(a_n{)}_{n\ge 1}$ if there exists a subsequence that converges to $\infty$ (or $-\infty$).

Example: Accumulation Points of $a_n=(-1{)}^n+\frac{1}{n}$

Consider the sequence $a_n=(-1{)}^n+\frac{1}{n}$. We identified earlier that ${\mathrm{liminf}}_{n\to \infty }{a}_n=-1$ and ${\mathrm{limsup}}_{n\to \infty }{a}_n=1$.

Let’s find subsequences converging to $-1$ and $1$:

• Subsequence converging to 1: Consider the subsequence of terms with even indices, $a_{2k}=(-1{)}^{2k}+\frac{1}{2k}=1+\frac{1}{2k}$. As $k\to \infty$, $a_{2k}\to 1$. Thus, 1 is an accumulation point.

• Subsequence converging to -1: Consider the subsequence of terms with odd indices, $a_{2k-1}=(-1{)}^{2k-1}+\frac{1}{2k-1}=-1+\frac{1}{2k-1}$. As $k\to \infty$, $a_{2k-1}\to -1$. Thus, -1 is an accumulation point.

In fact, for $a_n=(-1{)}^n+\frac{1}{n}$, the accumulation points are precisely $\{-1,1\}$.

Accumulation Points for the Enumeration of Rationals in $[0,1]$

For the sequence $(a_n{)}_{n\ge 1}$ that enumerates $\mathbb{Q}\cap [0,1]$, what is the set of all accumulation points?

Answer: Every real number in the interval $[0,1]$ is an accumulation point of $(a_n)$.

Justification: Let $c\in [0,1]$ be any real number. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for any $ϵ>0$, the interval $(c-ϵ,c+ϵ)$ contains infinitely many rational numbers. Thus, we can construct a subsequence of $(a_n)$ that converges to $c$. Start by picking a term $a_{n_1}$ in $(c-1,c+1)\cap [0,1]\cap \mathbb{Q}$. Then pick $a_{n_2}$ with $n_2>n_1$ in $(c-\frac{1}{2},c+\frac{1}{2})\cap [0,1]\cap \mathbb{Q}$, and so on. This subsequence $(a_{n_k})$ will converge to $c$.

Connection between Accumulation Points, Limit Inferior, and Limit Superior

Let $(a_n{)}_{n\ge 1}$ be a real sequence.

1. The limit inferior ${\mathrm{liminf}}_{n\to \infty }{a}_n$ is the smallest accumulation point of $(a_n)$. The limit superior ${\mathrm{limsup}}_{n\to \infty }{a}_n$ is the largest accumulation point of $(a_n)$.

2. A sequence $(a_n{)}_{n\ge 1}$ converges to a limit $L\in \mathbb{R}$ if and only if $L$ is the unique accumulation point of $(a_n{)}_{n\ge 1}$.

Sequences in ${\mathbb{R}}^d$ and $\mathbb{C}$

Definition of Sequences in ${\mathbb{R}}^d$

A sequence in ${\mathbb{R}}^d$ is a function $a:{\mathbb{N}}^{\ast }\to {\mathbb{R}}^d$. We write $a_n$ instead of $a(n)$ and denote the sequence as $(a_n{)}_{n\ge 1}$. Each term $a_n$ is a vector in ${\mathbb{R}}^d$.

Recall: Euclidean Norm in ${\mathbb{R}}^d$

For $x=(x_1,x_2,\dots ,x_d)\in {\mathbb{R}}^d$, the Euclidean norm is defined as: $||x||=||(x_1,\dots ,x_d)||=\sqrt{x_1^2+x_2^2+\cdots +x_d^2}=\sqrt{{\sum }_{j=1}^d{x}_j^2}$

Convergence of Sequences in ${\mathbb{R}}^d$

A sequence $(a_n{)}_{n\ge 1}$ in ${\mathbb{R}}^d$ is called convergent if there exists a vector $a\in {\mathbb{R}}^d$ such that for every $ϵ>0$, there exists $N\in \mathbb{N}$ such that for all $n>N$, $||a_n-a||<ϵ$.

If such an $a$ exists, it is uniquely determined and is called the limit (Grenzwert) of the sequence $(a_n)$. We write $a={\lim }_{n\to \infty }{a}_n$.

Cauchy Sequences in ${\mathbb{R}}^d$

A sequence $(a_n{)}_{n\ge 1}$ in ${\mathbb{R}}^d$ is called a Cauchy sequence if for every $ϵ>0$, there exists $N\in \mathbb{N}$ such that for all $m,n>N$, $||a_m-a_n||<ϵ$.

Bounded Sequences in ${\mathbb{R}}^d$

A sequence $(a_n{)}_{n\ge 1}$ in ${\mathbb{R}}^d$ is called bounded if there exists $R\ge 0$ such that for all $n\ge 1$, $||a_n||\le R$.

Component-wise Convergence in ${\mathbb{R}}^d$

Let $(a_n{)}_{n\ge 1}$ be a sequence in ${\mathbb{R}}^d$, where $a_n=(a_{n,1},a_{n,2},\dots ,a_{n,d})$. Let $b=(b_1,b_2,\dots ,b_d)\in {\mathbb{R}}^d$. Then:

• Convergence Equivalence: The sequence $(a_n{)}_{n\ge 1}$ converges to $b$ in ${\mathbb{R}}^d$ if and only if for each component $j\in \{1,2,\dots ,d\}$, the sequence of $j$-th components $(a_{n,j}{)}_{n\ge 1}$ converges to $b_j$ in $\mathbb{R}$.

  ${\lim }_{n\to \infty }{a}_n=b⟺{\lim }_{n\to \infty }{a}_{n,j}=b_j\text{ for all }j\in \{1,2,\dots ,d\}$

• Cauchy Sequence Equivalence: The sequence $(a_n{)}_{n\ge 1}$ is a Cauchy sequence in ${\mathbb{R}}^d$ if and only if for each component $j\in \{1,2,\dots ,d\}$, the sequence of $j$-th components $(a_{n,j}{)}_{n\ge 1}$ is a Cauchy sequence in $\mathbb{R}$.

• Boundedness Equivalence: The sequence $(a_n{)}_{n\ge 1}$ is bounded in ${\mathbb{R}}^d$ if and only if for each component $j\in \{1,2,\dots ,d\}$, the sequence of $j$-th components $(a_{n,j}{)}_{n\ge 1}$ is bounded in $\mathbb{R}$.

Proof Idea

These equivalences rely on the relationship between the norm in ${\mathbb{R}}^d$ and the components. For $x=(x_1,\dots ,x_d)\in {\mathbb{R}}^d$ and $1\le j\le d$:

$|x_j{|}^2\le {\sum }_{i=1}^d{x}_i^2=||x|{|}^2\le {\sum }_{i=1}^d({\max }_{1\le k\le d}|x_k|{)}^2=d\cdot ({\max }_{1\le k\le d}|x_k|{)}^2$

Taking square roots:

$|x_j|\le ||x||\le \sqrt{d}\cdot {\max }_{1\le k\le d}|x_k|\le \sqrt{d}\cdot {\sum }_{k=1}^d|x_k|$

This shows that convergence, Cauchy property, and boundedness in ${\mathbb{R}}^d$ are equivalent to the same properties holding component-wise in $\mathbb{R}$.

Detailed Implication for Convergence:

Let $a_n=(a_{n,1},\dots ,a_{n,d})$ and $b=(b_1,\dots ,b_d)$.

($\Rightarrow$) Assume ${\lim }_{n\to \infty }{a}_n=b$. Then for every $ϵ>0$, there is $N$ such that for $n>N$, $||a_n-b||<ϵ$. Since $|a_{n,j}-b_j|\le ||a_n-b||<ϵ$, we have ${\lim }_{n\to \infty }{a}_{n,j}=b_j$ for each $j$.

($\Leftarrow$) Assume ${\lim }_{n\to \infty }{a}_{n,j}=b_j$ for all $j$. For each $j$ and $ϵ>0$, there is $N_j$ such that for $n>N_j$, $|a_{n,j}-b_j|<\frac{ϵ}{\sqrt{d}}$. Let $N=\max \{N_1,\dots ,N_d\}$. For $n>N$, for all $j$, $|a_{n,j}-b_j|<\frac{ϵ}{\sqrt{d}}$. Then,

$||a_n-b||=\sqrt{{\sum }_{j=1}^d(a_{n,j}-b_j{)}^2}<\sqrt{{\sum }_{j=1}^d(\frac{ϵ}{\sqrt{d}}{)}^2}=\sqrt{{\sum }_{j=1}^d\frac{{ϵ}^2}{d}}=\sqrt{d\cdot \frac{{ϵ}^2}{d}}=ϵ$ Thus, ${\lim }_{n\to \infty }{a}_n=b$.

Convergence of Sequences in $\mathbb{C}$

Since $\mathbb{C}$ can be identified with ${\mathbb{R}}^2$, the concept of convergence in $\mathbb{C}$ is analogous to convergence in ${\mathbb{R}}^2$. For a complex sequence $(z_n{)}_{n\ge 1}$, where $z_n=x_n+i{y}_n$, and a complex number $c=a+ib$, we have:

${\lim }_{n\to \infty }{z}_n=c⟺{\lim }_{n\to \infty }\text{Re}(z_n)=a\text{ and }{\lim }_{n\to \infty }\text{Im}(z_n)=b$

Equivalently, ${\lim }_{n\to \infty }{z}_n=c$ if and only if ${\lim }_{n\to \infty }|z_n-c|=0$.

Algebraic Limit Laws for Complex Sequences:

The algebraic limit laws (sum, product, quotient) discussed for real sequences also hold for complex sequences. For example, if ${\lim }_{n\to \infty }{z}_n=z$ and ${\lim }_{n\to \infty }{w}_n=w$, then ${\lim }_{n\to \infty }(z_n+w_n)=z+w$ and ${\lim }_{n\to \infty }(z_n{w}_n)=zw$.

Example: Convergence of a Complex Sequence

Consider the complex sequence $z_n=\frac{(-1{)}^n}{n}+\frac{n+1}{n}i$.

To determine the limit of this sequence, we analyze the real and imaginary parts separately:

• Real Part: $\text{Re}(z_n)=\frac{(-1{)}^n}{n}$. As $n\to \infty$, the numerator $(-1{)}^n$ oscillates between -1 and 1, while the denominator $n$ grows infinitely large. Therefore, the real part converges to 0: ${\lim }_{n\to \infty }\text{Re}(z_n)={\lim }_{n\to \infty }\frac{(-1{)}^n}{n}=0$

• Imaginary Part: $\text{Im}(z_n)=\frac{n+1}{n}$. We can rewrite this as: $\text{Im}(z_n)=\frac{n+1}{n}=\frac{n}{n}+\frac{1}{n}=1+\frac{1}{n}$ As $n\to \infty$, $\frac{1}{n}\to 0$, so the imaginary part converges to 1: ${\lim }_{n\to \infty }\text{Im}(z_n)={\lim }_{n\to \infty }\left(1+\frac{1}{n}\right)=1$

Since both the real and imaginary parts converge, the complex sequence $(z_n)$ converges to the complex number formed by these limits:

${\lim }_{n\to \infty }{z}_n=\left({\lim }_{n\to \infty }\text{Re}(z_n)\right)+i\left({\lim }_{n\to \infty }\text{Im}(z_n)\right)=0+i\cdot 1=i$

Thus, ${\lim }_{n\to \infty }{z}_n=i$.

Introduction to Series (Reihen)

Definition of a Series

Let $(a_k{)}_{k\ge 1}$ be a sequence in $\mathbb{R}$ or $\mathbb{C}$. A series is formally denoted as ${\sum }_{k=1}^{\infty }{a}_k$. It is defined as the sequence of partial sums $(S_n{)}_{n\ge 1}$, where $S_n={\sum }_{k=1}^n{a}_k=a_1+a_2+\cdots +a_n$.

The series ${\sum }_{k=1}^{\infty }{a}_k$ is said to be convergent if the sequence of partial sums $(S_n{)}_{n\ge 1}$ converges. In this case, the value of the series is defined as:

${\sum }_{k=1}^{\infty }{a}_k={\lim }_{n\to \infty }{S}_n={\lim }_{n\to \infty }{\sum }_{k=1}^n{a}_k$

If the sequence $(S_n{)}_{n\ge 1}$ diverges, the series ${\sum }_{k=1}^{\infty }{a}_k$ is said to be divergent.

Geometric Series Example

Let $q\in \mathbb{C}$ with $|q|<1$. Consider the geometric series ${\sum }_{k=0}^{\infty }{q}^k=1+q+q^2+q^3+\dots$.

Let’s find the partial sums $S_n={\sum }_{k=0}^n{q}^k=1+q+q^2+\cdots +q^n$.

Multiply $S_n$ by $q$: $q{S}_n=q+q^2+\cdots +q^n+q^{n+1}$.

Subtract $q{S}_n$ from $S_n$: $(1-q)S_n=S_n-q{S}_n=(1+q+\cdots +q^n)-(q+q^2+\cdots +q^{n+1})=1-q^{n+1}$.

If $q\ne 1$, then $S_n=\frac{1-q^{n+1}}{1-q}$.

Since $|q|<1$, we know that ${\lim }_{n\to \infty }{q}^{n+1}=0$. Therefore,

${\sum }_{k=0}^{\infty }{q}^k={\lim }_{n\to \infty }{S}_n={\lim }_{n\to \infty }\frac{1-q^{n+1}}{1-q}=\frac{1-0}{1-q}=\frac{1}{1-q}$

Thus, the geometric series converges to $\frac{1}{1-q}$ when $|q|<1$.

Example Calculation: ${\sum }_{k=0}^{\infty }(\frac{1}{2}{)}^k=\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2$.

Divergent Harmonic Series Example

Consider the harmonic series ${\sum }_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$.

This series is divergent.

The partial sums increase by at least $\frac{1}{2}$ in each group, so they will grow without bound. The partial sums $(S_n)$ of the harmonic series are not convergent.

Algebraic Properties of Convergent Series

Let ${\sum }_{k=1}^{\infty }{a}_k$ and ${\sum }_{k=1}^{\infty }{b}_k$ be convergent series, and let $\alpha \in \mathbb{C}$ (or $\mathbb{R}$). Then:

1. The series ${\sum }_{k=1}^{\infty }(a_k+b_k)$ is convergent, and ${\sum }_{k=1}^{\infty }(a_k+b_k)={\sum }_{k=1}^{\infty }{a}_k+{\sum }_{k=1}^{\infty }{b}_k$

2. The series ${\sum }_{k=1}^{\infty }(\alpha a_k)$ is convergent, and ${\sum }_{k=1}^{\infty }(\alpha a_k)=\alpha {\sum }_{k=1}^{\infty }{a}_k$

Proof Sketch for (2)

Let $S_n={\sum }_{k=1}^n{a}_k$ be the partial sums of $\sum a_k$. Let $T_n={\sum }_{k=1}^n(\alpha a_k)$. Then $T_n=\alpha {\sum }_{k=1}^n{a}_k=\alpha S_n$. If ${\sum }_{k=1}^{\infty }{a}_k$ converges to $L={\lim }_{n\to \infty }{S}_n$, then ${\lim }_{n\to \infty }{T}_n={\lim }_{n\to \infty }(\alpha S_n)=\alpha {\lim }_{n\to \infty }{S}_n=\alpha L=\alpha {\sum }_{k=1}^{\infty }{a}_k$.

Series as Telescoping Sums

Question: Is every sequence a series?

Answer: Yes, every sequence can be represented as a series.

Let $(b_n{)}_{n\ge 1}$ be a sequence. Define $a_1=b_1$, and for $n\ge 2$, $a_n=b_n-b_{n-1}$. Then consider the series ${\sum }_{k=1}^{\infty }{a}_k$. The partial sums are:

$S_n={\sum }_{k=1}^n{a}_k=a_1+a_2+\cdots +a_n=b_1+(b_2-b_1)+(b_3-b_2)+\cdots +(b_n-b_{n-1})=b_n$.

This is a telescoping sum. Thus, the partial sums of ${\sum }_{k=1}^{\infty }{a}_k$ are exactly the terms of the sequence $(b_n{)}_{n\ge 1}$. Therefore, the series ${\sum }_{k=1}^{\infty }{a}_k$ converges if and only if the sequence $(b_n{)}_{n\ge 1}$ converges, and the value of the series is ${\lim }_{n\to \infty }{b}_n$.

Importance of Series: While every sequence can be seen as a series, the series perspective is often very useful in theory and applications. Infinite sums arise naturally in many areas, from ancient paradoxes like Zeno’s paradox to modern applications in electro-technology (Fourier series) and beyond. The theory of series provides powerful tools for analyzing and working with infinite sums.

Continue here: 07 Series Cauchy Criterion, Absolute and Conditional Convergence, Riemann Rearrangement Theorem, Dirichlet’s Theorem