17 minute read

We begin with limits of sequences.

Some of the concepts and results included in this chapter have been encountered in the study of calculus.

Our presentation will be considerably more rigorous - emphasizing proofs rather than computation.

1. Convergent Squences

[DEFINITION 3.1.1]
A Sequence $ {p_n }$ in $X$ is said to converge if there exists a point $p \in X$ such that for every $\epsilon > 0$, there exists a positvie integer $n_0 = n_0(\epsilon)$ such that $p_n \in N_\epsilon (p)$ for all $n \geq n_0$. If this is the case, we say that ${ p_n }$ converges to $p$.
In this is the case, we say that $p_n$ converges to $p$, or that $p$ is the limist of the sequence $p_n$, and we write $\lim_{n \rightarrow \infty} p_n = p$ or $p_n \rightarrow p$.
If $p_n$ does not converge, then $p_n$ is said to diverge.
In the definition, the statement $p_n \in N_{\epsilon}(p)$ for all $n \geq n_0$ is equivalent to $d(p_n, p) < \epsilon$ for all $n \geq n_0$.

[DEFINITION 3.1.3]
A sequence ${p_n}$ in X is said to be bounded if there exists $p \in X$ and a positvie constant $M$ such that $d(p, p_n) \leq M$ for all $n \in \mathbb{N}$.

앞서 집합의 bounded를 정의했다면, 수열의 bounded도 정의한다.

[THEOREM 3.1.4]
Let $(X, d)$ be a metric space.
(a) If a sequence $ {p_n}$ in $X$ converges, then its limit is unique.
(b) Every convergent sequence in $X$ is bounded.
(c) If $E \subset X$ and $p$ is a limit point of $E$, then there exists a sequence ${p_n}$ in $E$ with $p_n \neq p$ for all $n$ such that $\lim_{x\to\infty} {p_n} = p$.

Proof (a)

We will prove it using proof by contradiction.

Suppose that the sequence $p_n$ converges to two distinct points $p, q \in X$.

Let $\epsilon = \dfrac{1}{3} d(p, q)$.

Since $p_n \rightarrow p, \exists n_1 \in \mathbb{N} \; s.t. \; \vert p_n - p \vert < \epsilon \; \forall n \geq n_1.$

Also, since $p_n \rightarrow p, \exists n_2 \in \mathbb{N} \; s.t. \; \vert p_n - q \vert < \epsilon \; \forall n \geq n_2.$

Thus if $n \geq max(n_1, n_2)$, by triangule inequality, $d(p, q) \leq d(p_n, p) + d(p_n, q) < 2 \epsilon = \dfrac{2}{3} d(p, q)$, which is a contradiction. Thereofre, $p = q$ and the sequence $p_n$ converges to unique point.

Proof (b)

Let ${p_n}$ be a convergent sequence in $X$ such that converges to $p \in X$.

Take $\epsilon = 1$. For this $\epsilon, \exists n_0 \in \mathbb{N} \; s.t. \; d(p_n, p) < \epsilon \; \forall n > n_0$. Let $M = max(d(p, p_1), d(p, p_2), \dots, d(p, p_{n_0}), 1)$.

Then $d(p, p_n) \leq M \; \forall n \in \mathbb{N}$.

Therefore $p_n$ is bounded.

Proof (c)

2. Sequences of Real Numbers

We will emphasize some of the important properties of sequences of real numbers, and also investigate the limites of several basic sequences that are frequently encountered in the analysis.

[THEOREM 3.2.1]
If ${ a_n }$ and ${ b_n }$ are convergent sequences of real numbers with $\lim_{n \rightarrow \infty} a_n= a$ and $\lim_{n \rightarrow \infty} b_n= b$,
then
(a) $\lim_{n \rightarrow \infty} (a_n + b_n) = a + b$, and
(b) $\lim_{n \rightarrow \infty} a_nb_n= ab$.
(c) Futhermore, if $a \neq 0$, and $a_n \neq 0$ for all $n$, then $\lim_{x\to\infty} \dfrac{b_n}{a_n}= \dfrac{b}{a}$.

Proof (a)

Since $a_n \rightarrow a$ and $b_n \rightarrow b$, $\forall \epsilon_1 \; \exists n_1 \; s.t. \; \vert a_n - a \vert < \epsilon_1 \; \forall n \geq n_1$ and $\forall \epsilon_2 \; \exists n_2 \; s.t. \; \vert b_n - b \vert < \epsilon_2 \; \forall n \geq n_2$.

Take $\epsilon = \epsilon_1 + \epsilon_2$ and $n_0 = max(n_1, n_2)$. if $n \geq n_0$, using triangle inequality, $\vert (a_n + b_n) - (a + b) \vert \leq \vert a_n - a \vert + \vert b_n - b \vert \leq \epsilon_1 + \epsilon_2 = \epsilon$.

Therefore, $(a_n + b_n) \rightarrow (a+b)$.

Proof (b)

Since $a_n \rightarrow a$, by theorem 3.1.4 (b), $a_n$ is bounded. Thus $\exists M >0 \; s.t. \; \vert a_n \vert \leq M \; \forall \; n \in \mathbb{N}$.

Take $\epsilon = M \epsilon_2 + \vert b \vert \epsilon_1 > 0$ and $n_0 = max(n_1, n_2)$.

Since $a_n \rightarrow a$ and $b_n \rightarrow b$, $\forall \epsilon_1 > 0 \; \exists n_1 \in \mathbb{N} \; s.t. \; \vert a_n - a \vert < \epsilon_1 \; \forall n \geq n_1$ and $\forall \epsilon_2 > 0 \in \mathbb{N} \; \exists n_2 \; s.t. \; \vert b_n - b \vert < \epsilon_2 \; \forall n \geq n_2$.

If $n \geq n_0$,

\[\vert a_nb_n - ab \vert = \vert a_n (b_n - b) + b (a_n - a) \vert \\ \leq \vert a_n (b_n - b) \vert + \vert b (a_n - a) \vert \\ = \vert a_n \vert \cdot \vert (b_n - b) \vert + \vert b \vert \cdot \vert (a_n - a) \vert \\ \leq M \cdot \vert (b_n - b) \vert + \vert b \vert \cdot \vert (a_n - a) \vert \\ = M \epsilon_2 + \vert b \vert \epsilon_1 \\ = \epsilon\]

Therefore $a_nb_n \rightarrow ab$

Proof (C)

It is suffice to show that $\dfrac{1}{a_n} \rightarrow \dfrac{1}{a}$.

The result (c) then follows from (b).

Since $a \neq 0$ and $a_n \rightarrow a$, let $\epsilon’ = \dfrac{1}{2} \vert a \vert$ and $\exists n_0 \in \mathbb{N} \; s.t. \; \vert a_n - a \vert < \epsilon’ \; \forall n \geq n_0$.

Since $\vert a \vert \leq \vert a_n - a \vert + \vert a_n \vert < \dfrac{1}{2} \vert a \vert + \vert a_n \vert \; \forall \; n \geq n_0$, $\vert a_n \vert > \dfrac{1}{2} \vert a \vert \; \forall \; n \geq n_0$.

Take $\epsilon = \dfrac{1}{\vert a \vert}$, and if $n \geq n_0$,

\[\vert \dfrac{1}{a_n} - \dfrac{1}{a} \vert \leq \dfrac{\vert a_n - a \vert}{\vert a_n \vert \vert a \vert} < \dfrac{\dfrac{1}{2} \vert a \vert}{\dfrac{1}{2} \vert a \vert \vert a \vert} = \dfrac{1}{\vert a \vert} = \epsilon\]

Therefore $\dfrac{1}{a_n} \rightarrow \dfrac{1}{a}$

[COROLLARY 3.2.2]
If ${ a_n }$ is a convergent sequence of real numbers with $\lim_{n \rightarrow \infty} a_n= a$, then for any $c \in \mathbb{R}$,
(a) $\lim_{n \rightarrow \infty} (a_n + c)= a + c$, and
(b) $\lim_{n \rightarrow \infty} c\;a_n= c\;a$.

Proof.

If we define the sequence $c_n$ by $c_n = c$ for all $n \in \mathbb{N}$, then the conclusions follow by (a) and (b) of the previous theorem.

[THEOREM 3.2.3]
Let ${ a_n }$ and ${ b_n }$ be a sequences of real numbers. If ${ b_n }$ is bounded and $\lim_{n \rightarrow \infty} a_n= 0$, then $\lim_{n \rightarrow \infty} a_n b_n= 0$

Proof.

Since $b_n$ is bounded, $\exists M > 0 \; s.t. \; \vert b_n \vert < M \;\forall \; n \in \mathbb{N}$.

Let $\epsilon > 0$ be given.

Since $a_n \rightarrow 0$, $\exists n_0 = n_0(\epsilon) \in \mathbb{N} \; s.t. \; \vert a_n - 0 \vert < \dfrac{\epsilon}{M} \; \forall n \geq n_0$.

Thus, if $n \geq n_0$, $\vert a_nb_n - 0 \vert \leq \vert a_n \vert \vert b_n \vert \leq \dfrac{\epsilon}{M} \cdot M = \epsilon$.

Therefore, $a_nb_n \rightarrow 0$.

[THEOREM 3.2.4] (Squeeze theorem)
Suppose ${a_n}$, ${b_n}$, and ${c_n}$ are sequences of real numbers for which there exists $n_0 \in \mathbb{N}$ such that $a_n \leq b_n \leq c_n$ for all $n \in \mathbb{N}$, $n \geq n_0$, and that $\lim_{n \rightarrow \infty} a_n= \lim_{n \rightarrow \infty} b_n = L$.
Then the sequence ${b_n}$ converges and $\lim_{n \rightarrow \infty} b_n = L$

Squeeze theorem (sandwich theorem)을 이용해서 정말 다양한 수열의 극한값을 구한다.

Proof.

Suppose ${a_n}$, ${b_n}$, and ${c_n}$ are sequences of real numbers for which there exists $n_0 \in \mathbb{N}$ such that $a_n \leq b_n \leq c_n$ for all $n \in \mathbb{N}$, $n \geq n_0$, and that $\lim_{n \rightarrow \infty} a_n= \lim_{n \rightarrow \infty} b_n = L$.

Let $\epsilon > 0$ be given.

Since $a_n \rightarrow L$ and $b_n \rightarrow L$,

$\exists n_1 \in \mathbb{N} \; s.t. \; \vert a_n - L \vert < \epsilon \; \forall n \geq n_1$ and $\exists n_2 \in \mathbb{N} \; s.t. \; \vert c_n - L \vert < \epsilon \; \forall n \geq n_2$

Thus, $L - \epsilon < a_n < L + \epsilon$ and $L - \epsilon < c_n < L + \epsilon$.

Take $n_0 = max(n_1, n_2)$.

Since $L - \epsilon < a_n \leq b_n \leq c_n < L + \epsilon \; \forall n \geq n_0$, $\vert b_n - L \vert < \epsilon \; \forall n \geq n_0$.

Therefore $c_n \rightarrow L$.

Some special Sequences

We consider some special sequences of real numbers that occur frequently in the study of analysis.

[THEOREM 3.2.5] (Binomial Theorem)
For $a \in \mathbb{R}$, $n \in \mathbb{N},$
$(1 + a)^n = \sum_{k=0}^n {n \choose k} a^k = {n \choose 0} + {n \choose 1} a + \dots + {n \choose n} a^n$

Proof.

We will use Mathematical induction.

When $n = 1$, ${1 \choose 0} + {1 \choose 1} a = 1 + 1$, which means the given identity is true.

Suppose that the given identity is true, which is ${n \choose 0} + {n \choose 1} a + \dots + {n \choose n} a^n$.

\[(1 + a)^{n+1} = (1 + a) (1 + a)^n \\ = (1 + a) \{ {n \choose 0} + {n \choose 1} a + \dots + {n \choose n} a^n \} \\ = \{ {n \choose 0} + {n \choose 1} a + \dots + {n \choose n} a^n \} + \{ {n \choose 0} a + {n \choose 1} a^2 + \dots + {n \choose n} a^{n+1} \} \\ = {n \choose 0} + {n+1 \choose 1} a + \dots + {n+1 \choose n} a^{n+1} + {n \choose n} a^{n+1} \\ = {n+1 \choose 0} + {n+1 \choose 1} a + \dots + {n+1 \choose n} a^{n+1} + {n+1 \choose n+1} a^{n+1}\]

, which means the given identity is true when $n = n + 1$.

Therefore, by mathematical induction, the given identity is true.

[THEOREM 3.2.6]
(a) If $p > 0$, then $\lim_{n \rightarrow \infty} \dfrac{1}{n^p} = 0$.
(b) If $p > 0$, then $\lim_{n \rightarrow \infty} \sqrt[n]{p} = 1$.
(c) $\lim_{n \rightarrow \infty} \sqrt[n]{n} = 1$
(d) If $p > 1$ and $\alpha$ is real, then $\lim_{n \rightarrow \infty} \dfrac{n^\alpha}{p^n} = 0$.
(e) If $|p| < 1$, then $\lim_{n \rightarrow \infty} p^n = 0$.
(f) For all $p \in \mathbb{R}$, $\lim_{n \rightarrow \infty} \dfrac{p^n}{n!} = 0$.

(a)를 증명해야 (d)를 증명할수 있고, (d)를 증명해야 (e)를 증명할 수 있고, (e)를 증명해야 (f)를 증명할 수 있다. 해석학에서는 고등학교 수학부터 당연시 사용하던 (e)를 엄밀하게 증명한다.

Proof (a)

Let $\epsilon \in \mathbb{R}$ s.t. $\epsilon > 0 $ be given.

By Archimedian property, $\exists n_0$ s.t. $n_0 \epsilon^{1/p} > 1$ or $\epsilon > \dfrac{1}{n_0^p}$.

If $n \geq n_0$, $\vert \dfrac{1}{n^p} - 0 \vert \leq \vert \dfrac{1}{n_0^p} \vert < \epsilon$.

Therefore, $\dfrac{1}{n^p} \rightarrow 0$

Proof (b)

When $p = 1$, $\lim_{n \rightarrow \infty} \sqrt[n]{p} = \lim_{n \rightarrow \infty} 1 = 1$.

When $p < 1$, $\exists q = \dfrac{1}{p}$.

If is suffice to show that $\lim_{n \rightarrow \infty} \sqrt[n]{p} = 1$ for $p > 1$.

Since $p > 1$, $\sqrt[n]{p} > 1$ and $\sqrt[n]{p} = p^{1/n} = 1 + a_n$ for $n \in \mathbb{N}$ and some $a_n > 0$.

\[p = (1 + a_n)^n = {n \choose 0} + {n \choose 1} a_n + \dots + {n \choose n} a_n^n \geq na_n\]

Since $0 < a_n < \dfrac{p}{n}$ for all $n \in \mathbb{N}$ and $\dfrac{p}{n} \rightarrow 0$, by squeeze theorem, $a_n \rightarrow 0$.

Therefore, $\lim_{n \rightarrow \infty} \sqrt[n]{p} = \lim_{n \rightarrow \infty} (1 + a_n) = 1$.

Proof (c)

Let $x_n = \sqrt[n]{n} - 1 > 0$ for $n \in \mathbb{N}$.

By the binomial theorem,

\[n = (x_n + 1)^n = 1 + nx_n + \dfrac{n(n-1)}{2}x_n^2 + \dots + x_n^n \geq \dfrac{n(n-1)}{2}x_n^2\]

for all $n \geq 2$.

Since $0 < x_n < \sqrt{\dfrac{2}{n-1}}$ for all $n \geq 2$ and $\lim_{n \rightarrow \infty} \sqrt{\dfrac{2}{n-1}} = 0$, by squeeze theorem, $x_n \rightarrow 0$.

Therefore, $\lim_{n \rightarrow \infty} \sqrt[n]{n} = \lim_{n \rightarrow \infty} (1 + x_n) = 1$.

3. Monotone Sequences

수열이 monotone 하다는 것은 굉장히 강력한 조건이다.

In this section, we will consider monotone sequences of real numbers. One of the advantages of such sequences is that they will either converge tn $\mathbb{R}$, or diverge to $+\infty$ or $-\infty$.

[Definition 3.3.1]
A sequence $a_n$ of real numbers is said to be
(a) monotone increasing (or nondecreasing) if $a_n \leq a_{n+1}$ for all $n \in \mathbb{N}$;
(b) monotone decreasing (or nonincreasing) if $a_n \geq a_{n+1}$ for all $n \in \mathbb{N}$;
(c) monotone if it is either monotone increasing or monotone decreasing.

A sequence $a_n$ is strictly increasing if $a_n < a_{n+1}$ for all $n \in \mathbb{N}$. A sequence $a_n$ is strictly decreasing if $a_n > a_{n+1}$ for all $n \in \mathbb{N}$.

[Theorem 3.3.2] (Monotone Convergence Theorem)
If $a_n$ is monotone and bounded, then $a_n$ converges.

사실상 bounded above면 충분하다.

이 정리는 수열의 수렴성 뿐만 아니라 수렴값에 대한 정보도 제공한다. 수열의 supremum으로 수렴한다고 말한다.

집합의 supremum을 통해 극한값을 알 수 있다.

Nested Intervals Property

[Corollary 3.3.3] (Nested Intervals Property)
If $I_n$ is a sequence of closed and bounded intervals with $I_n \supset I_{n+1}$ for all $n \in \mathbb{N}$, then
$\cap_{n=1}^{\infty} I_n \neq \emptyset$.

Euler’s Number $\mathbb{e}$

굉장히 아름다운 부분이다.

[Example 3.3.5]
We consider in detain the very important sequence $t_n = (1 + \dfrac{1}{n})^n$.
We will show that the sequence $t_n$ is monotone increasing and bounded above, and thus has a limit.
The standard notation for this limit is $\mathbb{e} = \lim_{n \to \infty} (1 + \dfrac{1}{n})^n$.

Infinite Limits

먼저 무한대로 발산을 엄밀하게 정의한다. 정의역에서 존재성을 가지고 정의하면 진술이 편해진다.

[Definition 3.3.6]
Let $a_n$ be a sequence of real numbers.
We say that $a_n$ approaches infinity, or that $a_n$ diverges to $\infty$, denoted $a_n \rightarrow \infty$,
if for every positive real number $M$, there exists an integer $n_0 \in \mathbb{N}$ such that $a_n > M$ for all $n \geq n_0$.

발산의 정의가 수렴의 정의와 굉장히 유사하다.

for every $\epsilon > 0$, there exists a positvie integer $n_0 = n_0(\epsilon)$ such that $p_n \in N_\epsilon (p)$ for all $n \geq n_0$

[Theorem 3.3.7]
If $a_n$ is monotone increasing and not bounded above, then $a_n \rightarrow \infty$ as $n \rightarrow \infty$.

4. Subsequences and the Bolzano-Weierstrass Theorem

Supremum property와 동치이다.

In this section, we will consider subsequences and subsequential limits of a given sequence of real numbers. One of the key results of the section is that every bounded sequence of real numbers bas a convergent subsequence. This result, also known as the sequential version of the Bolzano Weierstrass theorem, is one of the fundemental results of real analysis.

[Definition 3.4.1]
Let $(X, d)$ be a metric space.
Given a sequence $p_n$ in $X$, consider a sequence $n_k$ of positive integers such that $n_1 < n_2 < n_3 < \cdots$.
Then the sequence $p_{n_k}$ is called a subsequence of the sequence $p_n$.

A point $p \in X$ is a subsequential limit of the sequence ${p_n}$ is there exists a subsequence ${p_{n_k}}$ of ${p_n}$ that converges to p. Also, given a sequence ${p_n}$ in $\mathbb{R}$, we say that $\infty$ is a subsequential limit of ${p_n}$ if there exists a subsequence ${p_{n_k}}$ so that ${p_{n_k}} \rightarrow \infty$ as $k \rightarrow \infty$. Similarly for $-\infty$.

[Theorem 3.4.3]
Let $(X, d)$ be a metric space and let ${p_n}$ be a sequence in $X$.
If ${p_n}$ converges to $p$, then every subsequence of ${p_n}$ also converges to $p$.

[Corollary 3.4.6] (Bolzano-Weierstrass)
Every bounded sequence in $\mathbb{R}$ has convergent subsequence.

We continus in this way to obtain a sequence of nested interval $I_1 \supset I_2 \supset I_3 \dots \supset I_k \supset \dots$ and a subsequence $x_{n_k} \in I_k$ for $k \in \mathbb{N}$.

집합 $I$를 두 개로 나눴을 때 무한한 집합을 택해서 계속해서 나눈다.

5. Limit Superior and Inferior of a Sequence

In this section, we define the limit superior and limit inferior of a sequence of real numbers.

These two limit operations are important because unlike the limit of a sequence, the limit superior and limit inferior of a sequence always exist.

The concepts of the limit superior and limit inferior will also be important in our study of both series of real numbers and power series.

Let $s_n$ be a sequence in $\mathbb{R}$.

\[a_k = \inf \{s_n : n \geq k\} \\ b_k = \sup \{s_n : n \geq k\}\]

[Definition 3.5.1]
Let $s_n$ be a sequence in $\mathbb{R}$.

The limit superior of $s_n$, denoted $\overline{\lim\limits_{n \to \infty}}$ or $\overline{\lim} s_n$ , is defined as
$\overline{\lim\limits_{n \to \infty}} s_n = \lim\limits_{k \to \infty} b_k = \inf\limits_{k \in N} \sup [s_n : n \geq k]$

The limit inferior of $s_n$, denoted $\lim\limits_{\overline{n\to 0}}s_n$ or $\underline{\lim} s_n$, is defined as
$\lim\limits_{\overline{n\to 0}} s_n = \lim\limits_{k \to \infty} a_k = \sup\limits_{k \in N} \inf [s_n : n \geq k]$

집합과 수열을 이어주는 개념

잡합의 개념이었던 sup를 수열의 limsup을 정의함으로써 연결된다.

$\overline{\lim\limits_{\overline{n\to 0}}} s_n $

[Theorem 3.5.3]
Let $s_n$ be a sequence in $\mathbb{R}$.
(a) Suppose $\overline{\lim\limits_{n \to \infty}} s_n \in \mathbb{R}$. Then $\beta = \overline{\lim\limits_{n \to \infty}} s_n$ if and only if for all $\epsilon > 0$
(i) there exists $n_0 \in \mathbb{N}$ such that $s_n < \beta + \epsilon $ for all $n \geq n_0$, and
(ii) given $n \in \mathbb{N}$, there exists $k \in \mathbb{N}$ with $k \geq n$ such that $s_k > \beta - \epsilon$.

(b) $\overline{\lim\limits_{n \to \infty}} s_n = \infty$ if and only if given $M$ and $n \in \mathbb{N}$, there exists $k \in \mathbb{N}$ with $k \geq n$ such that $s_k \geq M$.

(c) $\overline{\lim\limits_{n \to \infty}} s_n = -\infty$ if and only if $s_n \rightarrow -\infty$ as $n \rightarrow \infty$.

[Corollary 3.5.5]
$\varlimsup_{n \to \infty} s_n = \varliminf_{n \to \infty} s_n$ if and ony if $\lim_{n \to \infty}$ exists in $\mathbb{R} \cup (-\infty, \infty)$.

6 Cauchy Sequences

We consider a criterion that for sequences in $\mathbb{R}$ is sufficient to ensure convergence of the sequence.

[definition 3.6.1]
Let (X, d) be a matric space.
A sequence $p_n$ in $X$ is a Cauchy sequence if for every $\epsilon > 0$, there exists a positive integer $n_0$ such that $d(p_n, p_m) < \epsilon$ for all integers $n, m \geq n_0$.

[Theorem 3.6.2]
Let (X, d) be a metric space.
(a) Every convergent sequence in $X$ is a Cauchy sequence.
(b) Every Cauchy sequence is bounded.

[Theorem 3.6.4]
If $p_n$ is a Cauchy sequence in a metric space $X$ that has a convergent subsequence, then the sequence $p_n$ converges.

[Theorem 3.6.5]
Every Cauchy sequence of real numbers converges.

[Theorem 3.6.6]
A metric space (X, d) is said to be completet if every Cauchy sequence in $X$ converges to a point in $X$.

Contractive Sequences

[Definition 3.6.8]
A sequence p_n in a metric space (X, d) is contractive if there exists a real number $b$, 0 < b < 1, such that $d(p_{n+1}, p_n) \leq b d(p_n, p_{n-1})$ for all $n \in mathbb{N}, n \geq 2$.

7 Series of Real Numbers

If $(a_n)_{n=1}^\infty$ is a sequence in $\mathbb{R}$ and if $p, q \in \mathbb{N}$ with $p \geq q$, set

\[\sum_{p=1}^qa_k = a_p + a_{p+1} + \dots + a_q\]

[Definition 3.7.1]
Let $(a_n){n=1}^\infty$ be a sequence of real numbers. Let $(s_n){n=1}^\infty$ be the sequence obtained from $a_n$, where for each $n \in \mathbb{N}$, $s_n = \sum_{k=1}^na_n$. \

The sequence $s_n$ is called an infinite series, or series, and is denoted either as $\sum_{k=1}^\infty a_k$ or $a_1 + a_2 + \dots + a_n + \dots$.
For each $n \in \mathbb{N}$, $s_n$ is called the $n$th partial sum of the series and $a_n$ is called the $n$th term of the series.

The series $\sum_{k=1}^\infty a_k$ converges if and only if the sequence $s_n$ of $n$th partial sums converges in $\mathbb{R}$.
If $\lim_{n \rightarrow \infty} s_n = s$, then $s$ is called the sum of the series, and we write $s = \sum_{k=1}^\infty a_k$.
If the sequence $s_n$ diverges, then the series $\sum_{k=1}^\infty a_k$ is said to diverge.

The Cauchy Criterion

The Cauchy criterion provides necessary and sufficient conditions for the convergence of a series.

[Theorem 3.7.3] (Cauchy Criterion)
The series $\sum_{k=1}^\infty$ a_k converges if and only if given $\epsilon >0$, there exists a positive integer $n_0$ such that $\sum_{k=n+1}^m a_k < \epsilon$ for all $m > n \geq n_0$.

[Corollary 3.7.5]
If $\sum_{k=1}^\infty a_k$ converges, then $\lim_{k \rightarrow \infty} a_k = 0$.

[Theorem 3.7.6]
Suppose $a_k \geq 0$ for all $k \in \mathbb{N}$. Then $\sum_{k=1}^\infty a_k$ converges if and only if $s_n$ is bounded above.

Reference

Manfred - Introduction to Real Analysis : Ch03

증명에 관하여

  • 증명 과정에서 어떤 목적에 의해 어떤 수를 도입할 때는 그 수의 소속을 밝혀야 한다. 자연수, 정수, 실수 등
  • 증명의 두입부에서 필요한 수들을 정의하는 과정인 필수적이다.
  • 정리를 증명할 때는 가장 먼저 해야 할 것 : 정리를 가정과 결론으로 나누고 정리의 가정을 증명의 가정으로 그래로 취한다.
  • 수열의 극한을 구한다는 것은 epsilon에 대한 특정한 부등식을 작성한다는 것이다. 부등식을 어떻게 작성할지가 핵심이다. Archimedian property, binomial theorem 등을 이용할 수 있다. 하지만 squeeze theorem을 이용하면 epsilon에 대한 부등식이 아닌 부등식을 이용해서 극한값을 구할 수 있다.