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# 复分析代考_Complex analysis代考_MA8108 Cauchy sequences, completeness of R, C

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## 复分析代考_Complex analysis代考_Cauchy sequences, completeness of R, C

So far, to determine convergence of a sequence required knowing the limit. In this section we prove an alternate machinery for deciding that a sequence converges without knowing what its limit is. This is used for subsequential limits and in comparison theorems for convergence of series in the next chapter.

Definition 8.7.1. A sequence $\left{s_n\right}$ is Cauchy if for all $\epsilon>0$ there exists a positive real number $N$ such that for all integers $m, n>N,\left|s_n-s_m\right|<\epsilon$. Theorem 8.7.2. Every Cauchy sequence is bounded. Proof. Let $\left{s_n\right}$ be a Cauchy sequence. Thus for $\epsilon=1$ there exists a positive integer $N$ such that for all integers $m, n>N,\left|s_n-s_m\right|<1$. Then the set $\left{\left|s_1\right|,\left|s_2\right|, \ldots,\left|s_N\right|,\left|s_{N+1}\right|\right}$ is a finite and hence a bounded subset of $\mathbb{R}$. Let $M^{\prime}$ an upper bound of this set, and let $M=M^{\prime}+1$. It follows that for all $n=1, \ldots, N,\left|s_n\right|N,\left|s_n\right|=$ $\left|s_n-s_{N+1}+s_{N+1}\right| \leq\left|s_n-s_{N+1}\right|+\left|s_{N+1}\right|<1+M^{\prime}=M$. Thus $\left{s_n\right}$ is bounded by $M$. Theorem 8.7.3. Every convergent sequence is Cauchy.

Proof. Let $\left{s_n\right}$ be a convergent sequence. Let $L$ be the limit. Let $\epsilon>0$. Since $\lim s_n=L$, there exists a positive real number $N$ such that for all $n>N,\left|s_n-L\right|<\epsilon / 2$. Thus for all integers $m, n>N$,
$$\left|s_n-s_m\right|=\left|s_n-L+L-s_m\right| \leq\left|s_n-L\right|+\left|L-s_m\right|<\epsilon / 2+\epsilon / 2=\epsilon .$$

## 复分析代考_Complex analysis代考_Subsequences

Definition 8.8.1. A subsequence of an infinite sequence $\left{s_n\right}$ is an infinite sequence $\left{s_{k_1}, s_{k_2}, s_{k_3}, \ldots\right}$ where $1 \leq k_1<k_2<k_3<\cdots$ are integers. Notations for such a subsequence are:
$$\left{s_{k_n}\right}, \quad\left{s_{k_n}\right}_n, \quad\left{s_{k_n}\right}_{n \geq 1}^{\infty}, \quad\left{s_{k_n}\right}_{n \geq 1}, \quad\left{s_{k_n}\right}_{n \in \mathbb{N}^{+}}$$
Note that for all $n, k_n \geq n$.
Examples 8.8.2. Every sequence is a subsequence of itself. Sequences ${1 / 2 n},{1 / 3 n}$, ${1 /(2 n+1)}$ are subsequences of ${1 / n}$, and ${1},{-1},\left{(-1)^{n+1}\right}$ are subsequences of $\left{(-1)^n\right}$. The constant sequence ${1}$ is not a subsequence of ${1 / n}$, because the latter sequence does not have infinitely many terms equal to 1 . If $\left{s_n\right}$ is the sequential enumeration of $\mathbb{Q}^{+}$on page 246 , then ${1 / n},{n},\left{\frac{n}{n+1}\right},\left{\frac{1}{n^2}\right}$ are subsequences, but $\left{\frac{n}{n+2}\right}={1 / 3,1 / 2, \ldots}$ is not.

Theorem 8.8.3. A subsequence of a convergent sequence is convergent, with the same limit. A subsequence of a Cauchy sequence is Cauchy.

Proof. Let $\left{s_n\right}$ be a convergent sequence, with limit $L$, and let $\left{s_{k_n}\right}$ be a subsequence. Let $\epsilon>0$. By assumption there exists a positive number $N$ such that for all integers $n>N,\left|s_n-L\right|<\epsilon$. Since $n \leq k_n$, it follows that $\left|s_{k_n}-L\right|<\epsilon$. Thus $\left{s_{k_n}\right}$ converges. The proof of the second part is similar.

## 复分析代考Complex analysis代考_Cauchy sequences, completeness of R, C

$$\left|s_n-s_m\right|=\left|s_n-L+L-s_m\right| \leq\left|s_n-L\right|+\left|L-s_m\right|<\epsilon / 2+\epsilon / 2=\epsilon .$$

## MATLAB代写

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