### 数学代写|微积分代写Calculus代写|Monotonic Sequences

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## 数学代写|微积分代写Calculus代写|Monotonic Sequences

We have seen that a bounded sequence need not converge. However, we shall show that, if the terms $a_{n}$ of a bounded sequence $\left(a_{n}\right)$ either increase or decrease as $n$ increases, then the sequence does converge.
Definition 1.1.8 A sequence $\left(a_{n}\right)$ is said to be a
(1) monotonically increasing sequence if $a_{n} \leq a_{n+1}$ for all $n \in \mathbb{N}$;
(2) monotonically decreasing sequence if $a_{n} \geq a_{n+1}$ for all $n \in \mathbb{N}$;
(3) monotonic sequence if it is either monotonically increasing or monotonically decreasing.

If strict inequality occur in (1) (resp. (2)), then we say that the sequence is strictly increasing (resp. strictly decreasing).

A monotonically increasing (respectively, a monotonically decreasing) sequence is also called an increasing (respectively, a decreasing) sequence.
We may observe that:

1. A sequence $\left(a_{n}\right)$ is monotonically increasing and bounded above if and only if $\left(-a_{n}\right)$ is monotonically decreasing and bounded below.
2. Every monotonically increasing sequence is bounded below, and every monotonically decreasing sequence is bounded above.

Example 1.1.20 The following statements can be easily verified (verify) (Figs. $1.7$ and 1.8):
(i) $\left(a_{n}\right)$ with $a_{n}=n /(n+1)$ is monotonically increasing.
(ii) ( $\left.a_{n}\right)$ with $a_{n}=(n+1) / n$ is monotonically decreasing.
(iii) $\left(a_{n}\right)$ with $a_{n}=(-1)^{n} n /(n+1)$ is neither monotonically increasing nor monotonically decreasing.
Exercise 1.1.14 Let $\left(a_{n}\right)$ be a monotonic sequence. Prove the following.
(i) Suppose $\left(a_{n}\right)$ converges. Then it is bounded above by the limit if it is increasing, and bounded below by the limit if it is decreasing.
(ii) Suppose $\left(a_{n}\right)$ diverges. Then it diverges to either infinity or minus infinity depending on whether it is increasing or decreasing.

Note that a convergent sequence need not be monotonically increasing or monotonically decreasing. For example, the sequence $\left((-1)^{n} / n\right)$ is convergent, but it is neither monotonically increasing nor monotonically decreasing. However, we have the following theorem. It helps us to show the convergence of many of the standard sequences.

## 数学代写|微积分代写Calculus代写|Subsequences

There are divergent sequences which contain terms which form convergent sequences. For example, the sequence $\left(a_{n}\right)$ with $a_{n}=(-1)^{n}$ is divergent, but the sequences $\left(a_{2 n-1}\right)$ and $\left(a_{2 n}\right)$ are convergent. Those sequences extracted from a given sequence, retaining the order in which the terms occur, are called subsequences. More precisely, we have the following definition.

Definition 1.1.12 A subsequence of a sequence $\left(a_{n}\right)$ is a sequence of the form $\left(a_{k_{n}}\right)$, where $\left(k_{n}\right)$ is a strictly increasing sequence of positive integers.
A sequence $\left(b_{n}\right)$ is a subsequence of a sequence $\left(a_{n}\right)$ if and only if there is a strictly increasing sequence $\left(k_{n}\right)$ of positive integers such that $b_{n}=a_{k_{n}}$ for all $n \in \mathbb{N}$.
For example, given a sequence $\left(a_{n}\right)$, the sequences
$$\left(a_{2 n}\right), \quad\left(a_{2 n+1}\right), \quad\left(a_{n^{2}}\right), \quad\left(a_{2^{n}}\right)$$
are some of the subsequences of $\left(a_{n}\right)$. As concrete examples,

(1) $\left(\frac{1}{2 n}\right),\left(\frac{1}{2 n+1}\right),\left(\frac{1}{n^{2}}\right)$ and $\left(\frac{1}{2^{n}}\right)$ are subsequences of $\left(\frac{1}{n}\right)$;
(2) $\left(\frac{1}{n}\right)$ and $\left(\frac{n}{n+1}\right)$ are subsequences of $\left(1, \frac{1}{2}, \frac{1}{2}, \frac{2}{3}, \frac{1}{3}, \frac{3}{4}, \ldots\right)$.
(3) The sequence $\left(b_{n}\right)$ with $b_{n}=1$ for all $n \in \mathbb{N}$ and the sequence $\left(c_{n}\right)$ with $c_{n}=-1$ for all $n \in \mathbb{N}$ are subsequences of the sequence $\left((-1)^{n}\right)$.
(4) $(2 n),(2 n+1),\left(n^{2}\right)$ and $\left(2^{n}\right)$ are subsequences of the sequence $\left(a_{n}\right)$ with $a_{n}=n$ for all $n \in \mathbb{N}$.

In (1), the given sequence and all the subsequences listed are convergent; in (2), the subsequences are convergent, but the original sequence is not convergent; same is the case with the case in (3); in (4), the original sequence and all the subsequences listed are divergent. In fact, in (4), every subsequence of the given sequence has to diverge to infinity (verify).
However, we have the following result.

## 数学代写|微积分代写Calculus代写|Further Examples

Example 1.1.28 Let a sequence $\left(a_{n}\right)$ be defined iteratively as follows: $a_{1}=1$ and
$$a_{n+1}=\frac{2 a_{n}+3}{4}$$
$n \in \mathbb{N}$. We show that $\left(a_{n}\right)$ is monotonically increasing and bounded above.
Note that
$$a_{n+1}=\frac{2 a_{n}+3}{4}=\frac{a_{n}}{2}+\frac{3}{4} \geq a_{n} \quad \Longleftrightarrow \quad a_{n} \leq \frac{3}{2}$$
Thus it is enough to show that $a_{n} \leq 3 / 2$ for all $n \in \mathbb{N}$.

Clearly, $a_{1} \leq 3 / 2$. If $a_{n} \leq 3 / 2$, then $a_{n+1}=a_{n} / 2+3 / 4<3 / 4+3 / 4=3 / 2$. Thus, we have proved that $a_{n} \leq 3 / 2$ for all $n \in \mathbb{N}$. Hence, by Theorem $1.1 .9,\left(a_{n}\right)$ converges. Let its limit be $a$. Then taking limit on both sides of $a_{n+1}=\frac{2 a_{n}+3}{4}$ we have
$$a=\frac{2 a+3}{4} \text { i.e., } 4 a=2 a+3 \text { so that } a=\frac{3}{2}$$
Another solution: Since $a:=3 / 2$ satisfies
$$a=\frac{2 a+3}{4}$$
we obtain
$$a_{n+1}-a=\frac{2 a_{n}+3}{4}-\frac{2 a+3}{4}=\frac{1}{2}\left(a_{n}-a\right)$$
Thus, by Theorem $1.1 .6, a_{n} \rightarrow a=3 / 2$
Example 1.1.29 Let a sequence $\left(a_{n}\right)$ be defined as follows : $a_{1}=2$ and
$$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right)$$
for $n \in \mathbb{N}$. Note that, if the sequence converges, then its limit $a \geq 0$, and then
$$a=\frac{1}{2}\left(a+\frac{2}{a}\right)$$
so that $a=\sqrt{2}$.
Since $a_{1}=2$, one may try to show that $\left(a_{n}\right)$ is monotonically decreasing and bounded below.
Note that
$$a_{n+1}:=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right) \leq a_{n} \Longleftrightarrow a_{n}^{2} \geq 2, \quad \text { i.e., } \quad a_{n} \geq \sqrt{2}$$
and
\begin{aligned} a_{n+1}:=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right) \geq \sqrt{2} & \Longleftrightarrow a_{n}^{2}-2 \sqrt{2} a_{n}+2 \geq 0 \ & \Longleftrightarrow\left(a_{n}-\sqrt{2}\right)^{2} \geq 0 \end{aligned}

## 数学代写|微积分代写Calculus代写|Monotonic Sequences

(1) 单调递增序列，如果一个n≤一个n+1对所有人n∈ñ;
(2) 单调递减序列如果一个n≥一个n+1对所有人n∈ñ;
(3) 单调序列，如果它是单调递增或单调递减的。

1. 一个序列(一个n)单调递增且有界当且仅当(−一个n)是单调递减且有界的。
2. 每个单调递增的序列都有界，每个单调递减的序列都有界。

(i)(一个n)和一个n=n/(n+1)是单调递增的。
(二) (一个n)和一个n=(n+1)/n是单调递减的。
㈢(一个n)和一个n=(−1)nn/(n+1)既不是单调递增也不是单调递减。

（我想(一个n)收敛。然后，如果它在增加，则以限制为界，如果它在减少，则以限制为界。
(ii) 假设(一个n)分歧。然后根据它是增加还是减少，它会发散到无穷大或负无穷大。

## 数学代写|微积分代写Calculus代写|Subsequences

(一个2n),(一个2n+1),(一个n2),(一个2n)

(1) (12n),(12n+1),(1n2)和(12n)是的子序列(1n);
(2) (1n)和(nn+1)是的子序列(1,12,12,23,13,34,…).
(3) 序列(bn)和bn=1对所有人n∈ñ和序列(Cn)和Cn=−1对所有人n∈ñ是序列的子序列((−1)n).
(4) (2n),(2n+1),(n2)和(2n)是序列的子序列(一个n)和一个n=n对所有人n∈ñ.

## 数学代写|微积分代写Calculus代写|Further Examples

n∈ñ. 我们表明(一个n)是单调递增且有界的。

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## MATLAB代写

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