### 数学代写|微积分代写Calculus代写| Cauchy Criterion

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## 数学代写|微积分代写Calculus代写|Cauchy Criterion

In Theorem 1.1.14, we have given a sufficient condition for the convergence of a sequence $\left(a_{n}\right)$, namely, that if $\left(a_{n}\right)$ satisfies
$$\left|a_{n+2}-a_{n+1}\right| \leq r\left|a_{n+1}-a_{n}\right| \forall n \in \mathbb{N}$$
for some $r$ with $0m$. Thus, $\left|a_{n}-a_{m}\right|$ can be made arbitrarily small for all large enough $n, m \in \mathbb{N}$. Now, we show that any sequence $\left(a_{n}\right)$ such that $\left|a_{n}-a_{m}\right|$ can be made arbitrarily small for all large enough $n, m \in \mathbb{N}$ actually converges. First, let us formally define the requirement on the sequence.

Definition 1.1.13 A a sequence $\left(a_{n}\right)$ is said to be a Cauchy sequence ${ }^{3}$ if for every $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that
$$\left|a_{n}-a_{m}\right|<\varepsilon \quad \forall n, m \geq N .$$
We have already observed in Remark $1.1 .15$ that if $\left(a_{n}\right)$ converges, then it need not satisfy the assumption in Theorem 1.1.14. However, we have the following theorem.
Theorem 1.1.15 Every convergent sequence is a Cauchy sequence.

Proof Suppose $\left(a_{n}\right)$ converges to $a$. Let $\varepsilon>0$ be given. Then we know that there exists $N \in \mathbb{N}$ such that $\left|a_{n}-a\right|<\varepsilon / 2$ for all $n \geq N$. Hence, we have
$$\left|a_{n}-a_{m}\right| \leq\left|a_{n}-a\right|+\left|a-a_{m}\right|<\varepsilon \forall n, m \geq N$$
This completes the proof.
Now, we show that the converse of Theorem 1.1.15 is also true. The idea of the proof is akin to the idea used in the proof of Theorem 1.1.14, namely, we first show that $\left(a_{n}\right)$ is a bounded sequence, so that by Bolzano-Weierstrass theorem (Theorem $1.1 .13),\left(a_{n}\right)$ has a subsequence which converges to some $a$, and then show that $\left(a_{n}\right)$ itself converges to $a$.

Theorem 1.1.16 (Cauchy criterion) Every Cauchy sequence of real numbers converges.

Proof Let $\left(a_{n}\right)$ be a Cauchy sequence. Taking $\varepsilon=1$, there exists $N \in \mathbb{N}$ such that
$$\left|a_{n}-a_{m}\right|<1 \quad \forall n, m \geq N .$$ In particular, for all $n \geq N$, $$\left|a_{n}\right| \leq\left|\left(a_{n}-a_{N}\right)+a_{N}\right| \leq\left|a_{n}-a_{N}\right|+\left|a_{N}\right| \leq 1+\left|a_{N}\right| .$$ Therefore, $$\left|a_{n}\right| \leq \max \left\{\left|a_{1}\right|,\left|a_{2}\right|, \ldots,\left|a_{N}\right|, 1+\left|a_{N}\right|\right\}$$ Thus, $\left(a_{n}\right)$ is a bounded sequence. As already mentioned, by Bolzano-Weierstrass theorem (Theorem $1.1 .13),\left(a_{n}\right)$ has a subsequence $\left(a_{k_{n}}\right)$ which converges to some $a$. Now, let $\varepsilon>0$ be given. Then there exist positive integers $N_{1}, N_{2}$ such that
$$\left|a_{k_{n}}-a\right|<\varepsilon / 2 \quad \forall n \geq N_{1}, \quad\left|a_{n}-a_{k_{n}}\right|<\varepsilon / 2 \quad \forall n \geq N_{2} .$$
Therefore,
$$\left|a_{n}-a\right| \leq\left|a_{n}-a_{k_{n}}\right|+\left|a_{k_{n}}-a\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon$$
for all $n \geq N_{3}:=\max \left{N_{1}, N_{2}\right}$. Thus, $a_{n} \rightarrow a$. This completes the proof.

## 数学代写|微积分代写Calculus代写|Series of Real Numbers

In the last section we have come across sequences whose terms involve some other sequences. For example, we had sequences such as $\left(a_{n}\right)$ with
(i) $a_{n}=\frac{3}{10}+\frac{3}{10^{2}}+\cdots+\frac{3}{10^{n}}$,
(ii) $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$,
(iii) $a_{n}=1+\frac{1}{2^{2}}+\cdots+\frac{1}{n^{2}}$.
Recall that, the sequence in (i) and (iii) converge whereas the sequence in (ii) diverge. In (i), we have also seen that the sequence converges to $\frac{1}{3}$. Because of this we may represent the number $\frac{1}{3}$ as
$$\frac{1}{3}=\frac{3}{10}+\frac{3}{10^{2}}+\cdots$$
More generally, we may have a sequence $\left(a_{n}\right)$ of real numbers, and we may form a new sequence $\left(s_{n}\right)$ by defining its $n^{\text {th }}$ term as sum of the first $n$ terms of $\left(a_{n}\right)$, that is,
$$s_{n}=a_{1}+\cdots+a_{n} .$$
Then we may enquire whether this new sequence converges or not. In case this sequence $\left(s_{n}\right)$ converge, then we may write its limit as
$$a_{1}+a_{2}+\cdots$$
This expression has a special name!

Definition 1.2.1 A series of real numbers is an expression of the form
$$a_{1}+a_{2}+a_{3}+\ldots$$
or more compactly,
$$\sum_{n=1}^{\infty} a_{n}$$
where $\left(a_{n}\right)$ is a sequence of real numbers. The number $a_{n}$ is called the $n^{\text {th }}$ term of the series and the sum of the first $n$ terms of $\left(a_{n}\right)$, that is,
$$s_{n}:=a_{1}+\cdots+a_{n},$$
is called the $n^{\text {th }}$ partial sum of the series.
Remark 1.2.1 Some authors denote a series as the sequence $\left(a_{n}, s_{n}\right)$, where $s_{n}$ is $n^{\text {th }}$ partial sum of the sequence $\left(a_{n}\right)$.
Example 1.2.1 Following are some examples of series:
(i) $\frac{3}{10}+\frac{3}{10^{2}}+\cdots$,
(ii) $1+\frac{1}{2}+\frac{1}{3}+\cdots$,
(iii) $1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots$
Note that in (i), (ii), (iii) above the partial sums are
$$\sum_{k=1}^{n} \frac{3}{10^{k}}, \quad \sum_{k=1}^{n} \frac{1}{k}, \quad \sum_{k=1}^{n} \frac{1}{k^{2}}$$
respectively.

## 数学代写|微积分代写Calculus代写|Convergence and Divergence of Series

The following definition is on expected lines (Fig. 1.9):
Definition 1.2.2 A series $\sum_{n=1}^{\infty} a_{n}$ is said to be a convergent series if the corresponding sequence $\left(s_{n}\right)$ of partial sums converges. If $s_{n} \rightarrow s$, then we say that the series $\sum_{n=1}^{\infty} a_{n}$ converges to $s$, and $s$ is called the sum of the series, and we write this fact as
$$s=\sum_{n=1}^{\infty} a_{n}$$

Observe the following: Suppose $a_{n} \geq 0$ for all $n \in \mathbb{N}$. Then the sequence $\left(s_{n}\right)$ of partial sums of the series $\sum_{n=1}^{\infty} a_{n}$ is monotonically increasing. Hence, in this case, either $\left(s_{n}\right)$ converges or $s_{n} \rightarrow \infty$.
Example 1.2.2 Consider the three series given in Example 1.2.1, i.e.,
$$\sum_{n=1}^{\infty} \frac{3}{10^{n}}, \quad \sum_{n=1}^{\infty} \frac{1}{n}, \quad \sum_{n=1}^{\infty} \frac{1}{n^{2}}$$
Note that the partial sums of these series, say $\left(s_{n}^{(1)}\right),\left(s_{n}^{(2)}\right),\left(s_{n}^{(3)}\right)$ with
$$s_{n}^{(1)}:=\sum_{k=1}^{n} \frac{3}{10^{k}}, \quad s_{n}^{(2)}:=\sum_{k=1}^{n} \frac{1}{k}, \quad s_{n}^{(3)}:=\sum_{k=1}^{n} \frac{1}{k^{2}},$$
are the sequences considered in Examples 1.1.5,1.1.19, 1.1.24, respectively, and we have seen that $\left(s_{n}^{(1)}\right)$ and $\left(s_{n}^{(3)}\right)$ converge, whereas $\left(s_{n}^{(2)}\right)$ diverges. Thus, $\sum_{n=1}^{\infty} \frac{3}{10^{n}}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ are convergent series, whereas $\sum_{n=1}^{\infty} \frac{1}{n}$ is a divergent series.
Example 1.2.3 Consider the geometric series
$$1+q+q^{2}+\cdots$$
for $q \in \mathbb{R}$. We show that this series converges if and only if $|q|<1$ :
Note that
$$s_{n}=1+q+\cdots+q^{n-1}= \begin{cases}n & \text { if } q=1 \ \left(1-q^{n}\right) /(1-q) & \text { if } q \neq 1\end{cases}$$

## 数学代写|微积分代写Calculus代写|Cauchy Criterion

|一个n+2−一个n+1|≤r|一个n+1−一个n|∀n∈ñ

|一个n−一个米|<e∀n,米≥ñ.

|一个n−一个米|≤|一个n−一个|+|一个−一个米|<e∀n,米≥ñ

|一个n−一个米|<1∀n,米≥ñ.特别是，对于所有n≥ñ,

|一个n|≤|(一个n−一个ñ)+一个ñ|≤|一个n−一个ñ|+|一个ñ|≤1+|一个ñ|.所以，

|一个n|≤最大限度{|一个1|,|一个2|,…,|一个ñ|,1+|一个ñ|}因此，(一个n)是有界序列。如前所述，由 Bolzano-Weierstrass 定理 (Theorem1.1.13),(一个n)有一个子序列(一个ķn)收敛到一些一个. 现在，让e>0被给予。那么存在正整数ñ1,ñ2这样

|一个ķn−一个|<e/2∀n≥ñ1,|一个n−一个ķn|<e/2∀n≥ñ2.

|一个n−一个|≤|一个n−一个ķn|+|一个ķn−一个|<e/2+e/2=e

## 数学代写|微积分代写Calculus代写|Series of Real Numbers

（一）一个n=310+3102+⋯+310n,
(ii)一个n=1+12+⋯+1n,
(iii)一个n=1+122+⋯+1n2.

13=310+3102+⋯

sn=一个1+⋯+一个n.

∑n=1∞一个n

sn:=一个1+⋯+一个n,

(i)310+3102+⋯,
(ii)1+12+13+⋯,
(iii)1+122+132+⋯

∑ķ=1n310ķ,∑ķ=1n1ķ,∑ķ=1n1ķ2

## 数学代写|微积分代写Calculus代写|Convergence and Divergence of Series

s=∑n=1∞一个n

∑n=1∞310n,∑n=1∞1n,∑n=1∞1n2

sn(1):=∑ķ=1n310ķ,sn(2):=∑ķ=1n1ķ,sn(3):=∑ķ=1n1ķ2,

1+q+q2+⋯

sn=1+q+⋯+qn−1={n 如果 q=1 (1−qn)/(1−q) 如果 q≠1

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