# 数学代写|实分析作业代写Real analysis代考|MAST20026

## 数学代写|实分析作业代写Real analysis代考|Bounded, unbounded, and divergent sequences

In this section, we briefly discuss sequences that don’t converge. A sequence that can be contained in a bounded interval is a bounded sequence.

For instance, the sequence in (3.5) is unbounded because it can’t be bounded by two real numbers, while the sequence in (3.4) is clearly bounded. The latter is actually a bounded sequence that doesn’t converge. On the other hand, every convergent sequence of real numbers is bounded since it approaches a limit and in particular, isn’t going far away. Here is how to prove this fact.
Theorem 60 Every convergent sequence of real numbers is bounded.
Proof Suppose $s_n$ is a convergent sequence with limit $s$. Then in particular with $\varepsilon=1$, there is a positive integer $N$ such that
$$\left|s_n-s\right|<1 \quad \text { for all } n>N$$
Therefore,
$$s-1N$$
So all terms $s_{N+1}, s_{N+2}, \ldots$ with indices greater than $N$ are bounded between the real number $s-1$ and $s+1$. That leaves a finite number of terms $s_1, s_2, \ldots, s_N$, which can’t get infinitely large. In fact, define
$$M=\max \left{\left|s_1\right|,\left|s_2\right|, \ldots,\left|s_N\right|\right}$$
and note that
$$-M \leq s_n \leq M$$
for $n=1,2, \ldots, N$. It follows that all terms of the sequence are within a bounded interval; we can define this interval in many ways, e.g.,
$$-M+s-1 \leq s_n \leq M+s+1 \text { for all } n \in \mathbb{N}$$
We now consider sequences that are not bounded or not convergent.

## 数学代写|实分析作业代写Real analysis代考|Subsequences

Consider the periodic sequence in (3.4)
$$s_n=\cos \frac{\pi n}{3}$$
Notice that for $n=6 k$ with $k=1,2,3, \ldots$ or $n=6,12,18, \ldots$ we get a sequence of constants: $s_{6 k}=\cos (2 \pi k)=1$. The sequence $s_{6 k}$ is a part of the original $s_n$ but unlike the latter, $s_{6 k}$ converges (trivially) to 1 as a constant sequence. You can check that there are five other parts of $s_n$ that also converge as constant sequences; they are $s_{6 k-1}, s_{6 k-2}, s_{6 k-3}, s_{6 k-4}$, and $s_{6 k-5}$.

Each of the six parts of the sequence $s_n$ above is called a subsequence of $s_n$. Although a nontrivial periodic sequence of period $m$ doesn’t converge, it has $m$ convergent subsequences. More generally, other types of non-convergent sequences can also have convergent subsequences, although the indices of the latter may not be evenly spaced. For instance, if
$$s_n=\cos (\pi \sqrt{n})$$
then $s_{4 k^2}=\cos (2 \pi k)=1$ is a convergent (constant) subsequence for $k=1,2,3, \ldots$ or $n=4 k^2=4,16,36, \ldots$ A less transparent example is $\sin n$ that has convergent subsequences; for this sequence, we need more powerful tools that are discussed in Chapter 4.

These results suggest that the concept of subsequence is important and useful in analysis, which is concerned with convergence and limits. We now give a formal definition.

# 实分析代考

## 数学代写|实分析作业代写Real analysis代考|Bounded, unbounded, and divergent sequences

$$\left|s_n-s\right|<1 \quad \text { for all } n>N$$

$$s-1 N$$

$$-M \leq s_n \leq M$$

$$-M+s-1 \leq s_n \leq M+s+1 \text { for all } n \in \mathbb{N}$$

## 数学代写|实分析作业代写Real analysis代考|Subsequences

$$s_n=\cos \frac{\pi n}{3}$$

$s_{6 k}=\cos (2 \pi k)=1$. 序列 $s_{6 k}$ 是原件的一部分 $s_n$ 但与后者不同的是， $s_{6 k}$ 作为常数序列收敛（平凡地） 到 1 。您可以检亘还有其他五个部分 $s_n$ 也收敛为常量序列；他们是 $s_{6 k-1}, s_{6 k-2}, s_{6 k-3}, s_{6 k-4}$ ， 和 $s_{6 k-5}$.

$$s_n=\cos (\pi \sqrt{n})$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: