text: # 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 $s_{6 k-5}$.
  
  $$s_n=\cos (\pi \sqrt{n})$$
  
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