### 数学代写|微积分代写Calculus代写|MATH141

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Monotone sequences

Definition 1.2.5. We say a sequence $\left{a_n\right}$ is monotone increasing if $a_n \leq a_{n+1}$ for all $n$. We say a sequence $\left{a_n\right}$ is monotone decreasing if $a_n \geq a_{n+1}$ for all $n$. We say a sequence is monotone if it is either monotone increasing or monotone decreasing.

Now suppose $\left{a_n\right}$ is a monotone increasing sequence. For such a sequence there either exists a number $P$ such that $a_n \leq P$ for all $n$ or there does not exist such a $P$. In the latter case, given any real number $M$, it is then possible to find an integer $N$ such that $a_N>M$. Since the sequence is monotone, it follows that $a_n>M$ for all $n>N$, and so the sequence diverges to infinity. On the other hand, if there does exist a number $P$ such that $a_n \leq P$ for all $n$, then there in fact exists a number $B$ such that $u_n \leq B$ for all $n$ and $B \leq P$ for any number $P$ with the property that $a_n \leq P$ for all $n$. The existence of $B$, known as the least upper bound of the sequence $\left{a_n\right}$, is not at all obvious; indeed, the subtle properties of the real numbers that imply the existence of $B$ were not fully understood until the middle part of the 19th century. However, given the existence of $B$, it is easy to see that given any $\epsilon>0$, there exists an integer $N$ for which $a_N>B-\epsilon$

(if not, then $B-\epsilon$ would be an upper bound for the sequence smaller than $B$ ). Since the sequence is monotone increasing and $a_nN$. That is, we have shown that the sequence converges and
$$\lim {n \rightarrow \infty} a_n=B \text {. }$$ Similar results hold for sequences which are monotone decreasing. Theurem 1.2.11 (Munutune sequence theurem). Suppose the sequence $\left{a_n\right}$ is monutone. If the sequence is monotone increasing and there exists a number $P$ such that $a_n \leq P$ for all $n$, then the sequence converges. If the sequence is monotone increasing and no such number $P$ exists, then $$\lim {n \rightarrow \infty} a_n=\infty .$$
If the sequence is monotone decreasing and there exists a number $Q$ such that $a_n \geq Q$ for all $n$, then the sequence converges. If the sequence is monotone decreasing and no such number $Q$ exists, then
$$\lim _{n \rightarrow \infty} a_n=-\infty$$

## 数学代写|微积分代写Calculus代写|The sum of a sequence

This section considers the problem of adding together the terms of a sequence. Of course, this is a problem only if more than a finite number of terms of the sequence are nonzero. In this case, we must decide what it means to add together an infinite number of nonzero numbers. The first example shows how a relatively simple question may lead to such infinite summations.

Example 1.3.1. Suppose a game is played in which a fair coin is tossed until the first time a head appears. What is the probability that a head appears for the first time on an even-numbered toss? To solve this problem, we first need to determine the probability of obtaining a head for the first time on any given even-numbered toss, and then we need to add all these probabilities together. Let $P_n$ denote the probability that the first head appears on the $n$th toss, $n=1,2,3, \ldots$. Then, since the coin is assumed to be fair,
$$P_1=\frac{1}{2} .$$
Now in order to get a head for the first time on the second toss, we must toss a tail on the first toss and then follow that with a head on the second toss. Since one-half of all first tosses will be tails and then one-half of those tosses will be followed by a second toss of heads, we should have
$$P_2=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4} .$$
Similarly, since one-fourth of all sequences of coin tosses will begin with two tails and then half of these sequences will have a head for the third toss, we have
$$P_3=\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)=\frac{1}{8} .$$
Continuing in this fashion, it should seem reasonable that, for any $n=1,2,3, \ldots$,
$$P_n=\frac{1}{2^n} .$$
Hence we have a sequence of probabilities $\left{P_n\right}$ for $n=1,2,3, \ldots$, and, in order to find the desired probability, we need to add up the even-numbered terms in this sequence. Namely, the probability that a head appears for the first time on an even toss is given by
$$P_2+P_4+P_6+\cdots=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots .$$

## 数学代写|微积分代写Calculus代写|Monotone sequences

$\backslash$ 左{a_n\右 $}$ 是单调递减的如果 $a_n \geq a_{n+1}$ 对所有人 $n$. 如果一个序列是单调递增或单调递减，我们就说它 是单调的。

(如果没有，那么 $B-\epsilon$ 将是小于的序列的上限 $B$ ). 由于序列是单调递增的，并且 $a_n N$. 也就是说，我们 已经证明序列收敛并且
$$\lim n \rightarrow \infty a_n=B .$$

$$\lim n \rightarrow \infty a_n=\infty .$$

$$\lim _{n \rightarrow \infty} a_n=-\infty$$

## 数学代写|微积分代写Calculus代写|The sum of a sequence

$$P_1=\frac{1}{2} .$$

$$P_2=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4} .$$

$$P_3=\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)=\frac{1}{8} .$$

$$P_n=\frac{1}{2^n} .$$

$$P_2+P_4+P_6+\cdots=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots$$

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