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• (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|Nonnegative Series and Decimal Expansions

Let $\left(a_n\right)$ be a sequence of reals. The series formed from the sequence $\left(a_n\right)$ is the sequence $\left(s_n\right)$ with terms $s_1=a_1, s_2=a_1+a_2$, and, for any $n \geq 1$, $s_n=a_1+a_2+\cdots+a_n$. The sequence $\left(s_n\right)$ is the sequence of partial sums. The terms $a_n$ are called summands, and the series is nonnegative if $a_n \geq 0$ for all $n \geq 1$. We often use sigma notation, and write $s_n=\sum_{k=1}^n a_k$. Series are often written
$$a_1+a_2+\ldots$$
In sigma notation, $\sum a_n$ or $\sum_{n=1}^{\infty} a_n$. If the sequence of partial sums $\left(s_n\right)$ has a limit $L$, then, we say the series sums or converges to $L$, and we write
$$L=a_1+a_2+\cdots=\sum_{n=1}^{\infty} a_n .$$
Then, $L$ is the sum of the series. By convention, we do not allow $\pm \infty$ as limits for series, only reals. Nevertheless, for nonnegative series, we write $\sum a_n=\infty$ to mean $\sum a_n$ diverges and $\sum a_n<\infty$ to mean $\sum a_n$ converges. As with sequences, sometimes it is more convenient to start a series from $n=0$. In this case, we write $\sum_{n=0}^{\infty} a_n$.

Let $L=\sum_{n=1}^{\infty} a_n$ be a convergent series and let $s_n$ denote its $n$th partial sum. The $n$th tail of the series is $L-s_n=\sum_{k=n+1}^{\infty} a_k$. Since the $n$th tail is the difference between the $n$th partial sum and the sum, we see that the nth tail of a convergent series goes to zero:
$$\lim {n / \infty} \sum{k=n+1}^{\infty} a_k=0 .$$
Let $a$ be real. Our first series is the geometric series
$$1+a+a^2+\cdots=\sum_{n=0}^{\infty} a^n .$$
Here the $n$th partial sum $s_n=1+a+\cdots+a^n$ is computed as follows:
$$a s_n=a\left(1+a+\cdots+a^n\right)=a+a^2+\cdots+a^{n+1}=s_n+a^{n+1}-1 .$$
Hence,
$$s_n=\frac{1-a^{n+1}}{1-a}, \quad a \neq 1$$

## 数学代写|微积分代写Calculus代写|Signed Series and Cauchy Sequences

A series is signed if its first term is positive, and at least one of its terms is negative. A series is alternating if it is of the form
$$\sum_{n=1}^{\infty}(-1)^{n-1} a_n=a_1-a_2+a_3 \cdots+(-1)^{n-1} a_n+\ldots$$ with $a_n$ positive for all $n \geq 1$. Alternating series are particularly tractable, but, first, we need a new concept.

A sequence (not a series!) $\left(a_n\right)$ is Cauchy if its terms approach each other, i.e., if there is a positive sequence $\left(e_n\right)$ converging to zero, such that
$$\left|a_{n+m}-a_n\right| \leq e_n, \quad \text { for all } m, n \geq 1 .$$
If a sequence is Cauchy, there are many choices for $\left(e_n\right)$. Any such sequence $\left(e_n\right)$ is an error sequence for the Cauchy sequence $\left(a_n\right)$.

It follows from the definition that every Cauchy sequence is bounded, $\left|a_m\right| \leq\left|a_m-a_1\right|+\left|a_1\right| \leq e_1+\left|a_1\right|$ for all $m \geq 1$.

It is easy to see that a convergent sequence is Cauchy. Indeed, if $\left(a_n\right)$ converges to $L$, then, $b_n=\left|a_n-L\right| \rightarrow 0$, so (S1.5), $b_n^* \rightarrow 0$. Hence, by the triangle inequality
$$\left|a_{n+m}-a_n\right| \leq\left|a_{n+m}-L\right|+\left|a_n-L\right| \leq b_n^+b_n^, \quad m>0, n \geq 1 .$$
Since $2 b_n^* \rightarrow 0,\left(2 b_n^*\right)$ is an error sequence for $\left(a_n\right)$, so, $\left(a_n\right)$ is Cauchy.
The following theorem shows that if the terms of a sequence “approach each other”, then, they “approach something”. To see that this is not a selfevident assertion, consider the following example. Let $a_n$ be the rational given by the first $n$ places in the decimal expansion of $\sqrt{2}$. Then, $\left|a_n-\sqrt{2}\right| \leq 10^{-n}$, hence, $a_n \rightarrow \sqrt{2}$, hence, $\left(a_n\right)$ is Cauchy. But, as far as $\mathbf{Q}$ is concerned, there is no limit, since $\sqrt{2} \notin \mathbf{Q}$. In other words, to actually establish the existence of the limit, one needs an additional property not enjoyed by $\mathbf{Q}$, the completeness property of $\mathbf{R}$.

# 微积分代考

## 数学代写|微积分代写calculus代写|非负级数和小数展开

. . > .

$$a_1+a_2+\ldots$$

$$L=a_1+a_2+\cdots=\sum_{n=1}^{\infty} a_n .$$

$$\lim {n / \infty} \sum{k=n+1}^{\infty} a_k=0 .$$

$$1+a+a^2+\cdots=\sum_{n=0}^{\infty} a^n .$$

$$a s_n=a\left(1+a+\cdots+a^n\right)=a+a^2+\cdots+a^{n+1}=s_n+a^{n+1}-1 .$$

$$s_n=\frac{1-a^{n+1}}{1-a}, \quad a \neq 1$$

## 数学代写|微积分代写calculus代写|Signed Series and Cauchy Sequences

.

$$\sum_{n=1}^{\infty}(-1)^{n-1} a_n=a_1-a_2+a_3 \cdots+(-1)^{n-1} a_n+\ldots$$, $a_n$为所有$n \geq 1$的正数，则该序列是交替的。交替级数特别容易处理，但是，首先，我们需要一个新概念

$$\left|a_{n+m}-a_n\right| \leq e_n, \quad \text { for all } m, n \geq 1 .$$

$$\left|a_{n+m}-a_n\right| \leq\left|a_{n+m}-L\right|+\left|a_n-L\right| \leq b_n^+b_n^, \quad m>0, n \geq 1 .$$

. . . . . . . .

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