数学代写|Math1131 Calculus

Let a sequence $\left(a_n\right)$ be defined iteratively as follows :
$$
a_1=1, \quad a_{n+1}=\frac{2+a_n}{1+a_n}, \quad n=1,2, \ldots
$$
Let us assume for a moment that $\left(a_n\right)$ converges to $a \in \mathbb{R}$. Then we obtain
$$
a=\frac{2+a}{1+a}
$$
i.e., $a(1+a)=2+a$, i.e., $a^2=2$. Thus, if $\left(a_n\right)$ converges, then the limit must be $a=\sqrt{2}$. Now, we prove that $\left(a_n\right)$ actually converges to $a:=\sqrt{2}$. Note that
$$
\begin{aligned}
a_{n+1}-a & =\frac{2+a_n}{1+a_n}-\frac{2+a}{1+a} \
& =\frac{a-a_n}{\left(1+a_n\right)(1+a)} .
\end{aligned}
$$
Note that $a_n \geq 1$ for all $n \in \mathbb{N}$ and $a \geq 1$. Hence, we obtain
$$
\left|a_{n+1}-a\right| \leq \frac{\left|a-a_n\right|}{\left(1+a_n\right)(1+a)} \leq \frac{\left|a_n-a\right|}{4} \quad \forall n \in \mathbb{N} .
$$
Therefore, by Theorem 1.1.6, $a_n \rightarrow a=\sqrt{2}$.

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.

Exercise 1.1 .15 Verify the above statements.
Before going further, let us illustrate the above notions by a few simple examples.
Example 1.1.21 We may observe the following:
(i) Intervals $(a, b),[a, b),(a, b],[a, b]$ with $a, b \in \mathbb{R}$ and $a<b$ are bounded sets.
(ii) For $a, b \in \mathbb{R}$, the intervals $(-\infty, b),(-\infty, b]$ are bounded above, but not bounded below, and the intervals $(a, \infty),[a, \infty)$ are bounded below but not bounded above.
(iii) The set $\mathbb{N}$ is bounded below, but not bounded above.
(iv) The sets $\mathbb{Z}$ of integers are neither bounded above nor bounded below.
(v) The set $\left{(-1)^n n: n \in \mathbb{N}\right}$ is neither bounded above nor bounded below.
We may also observe:

1. If $S \subseteq \mathbb{R}$ is bounded above (respectively, bounded below), then any subset of $S$ is bounded above (respectively, bounded below).
2. If $S \subseteq \mathbb{R}$ is not bounded above (respectively, not bounded below), then any super set of $S$ in $\mathbb{R}$ is not bounded above (respectively, not bounded below).

Exercise 1.1 .16 Verify the above statements.
Let $S \subseteq \mathbb{R}$. If $S$ is bounded above and if $b \in \mathbb{R}$ is an upper bound of $S$, then any number greater than $b$ is also an upper bound. So, one may seek a least of the upper bounds.

Definition 1.1.10 Let $S$ be a subset of $\mathbb{R}$. If $S$ is bounded above, and if $b_0$ is an upper bound of $S$ such that $b_0 \leq b$ for every upper bound $b$ of $S$, then $b_0$ is called the least upper bound or supremum for $S$, and it is denoted by $\operatorname{lub}(S)$ or $\sup (S)$.

Given any set of numbers having a certain property, it is not obvious whether there is a least number having the same property. For instance, if we consider the set
$$
{x \in \mathbb{Q}: 0<x<1}
$$
does not have any least number. However, the set of all upper bounds for a given set $S \subseteq \mathbb{R}$, does have a least of the upper bound, provided $S$ is non-empty. This is, in fact, one of the important properties of the set of all real numbers, called the least upper bound property (cf. Pugh [12]):

Least upper bound property: Every non-empty subset of $\mathbb{R}$ which is bounded above has the least upper bound.
Analogously, if $S$ is bounded below with a lower bound $a \in \mathbb{R}$, then any number less than $a$ is also a lower bound. In this case, we define the greatest lower bound as follows.