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# 数学代写|实分析代写Real Analysis代考|Series of positive terms

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## 数学代写|实分析代写Real Analysis代考|Series of positive terms

A series $\Sigma u_n$ is said to be a series of positive terms if $u_n$ is a positive real number for all $n \in \mathbb{N}$.

Theorem 6.2.1. A series of positive real numbers $\Sigma u_n$ is convergent if and only if the sequence $\left{s_n\right}$ of partial sums is bounded above.

Proof. $s_n=u_1+u_2+\cdots+u_n$. Then $s_{n+1}-s_n=u_{n+1}>0$ for all $n \in \mathbb{N}$.
Hence the sequence $\left{s_n\right}$ is a monotone increasing sqeuence. Therefore $\left{s_n\right}$ is convergent if and only if it bounded above.

Consequently, the series $\Sigma u_n$ is convergent if and only if the sequence $\left{s_n\right}$ is bounded above.

Note. If not bounded above, the sequence $\left{s_n\right}$ being a monotone increasing sequence, diverges to $\infty$. In this case the series diverges to $\infty$
Therefore a series of positive real numbers either converges to a real number, or diverges to $\infty$.

Introduction and removal of brackets.
Let $\Sigma u_n$ be a series of positive real numbers. Let the terms of the series be arranged in groups without changing the order of the terms. Let us denote the $n$th group by $v_n$. Then a new series $\Sigma v_n$ is obtained.

## 数学代写|实分析代写Real Analysis代考|Re-arrangement of terms

Let $\Sigma u_n$ be a given series. If a new series $\Sigma v_n$ is obtained by using each term of $\Sigma u_n$ exactly once, the order of the terms being disturbed, then $\Sigma v_n$ is called a re-arrangement of $\Sigma u_n$.

If $f: \mathbb{N} \rightarrow \mathbb{N}$ be a bijective mapping, $\Sigma u_{f(n)}$ is a re-arrangement of $\Sigma u_n$ and conversely if $\Sigma v_n$ be a re-arrangement of the series $\Sigma u_n$ then $v_n=u_{f(n)}$ for some bijection $f: \mathbb{N} \rightarrow \mathbb{N}$.
\begin{aligned} & \text { For example, let } f(n)=n+1 \text { if } n \text { be odd, } \ & =n-1 \text { if } n \text { be even. } \ & f(1)=2, f(2)=1, f(3)=4, f(4)=3, \cdots \cdots \ & \Sigma u_{f(n)}=u_2+u_1+u_4+u_3+\cdots \cdots \text { is a re-arrangement of } \Sigma u_n . \end{aligned}
Theorem 6.2.3. Let $\Sigma u_n$ be a convergent series of positive real numbers. Then any re-arrangement of $\Sigma u_n$ is convergent and the sum remains unaltered.

Proof. Let $\Sigma u_n$ converge to $s$ and $\Sigma v_n$ be a re-arrangement of $\Sigma u_n$. Then $v_n=u_{f(n)}$ for some bijection $f: \mathbb{N} \rightarrow \mathbb{N}$.
Let $s_n=u_1+u_2+\cdots+u_n, t_n=v_1+v_2+\cdots+v_n$.
Since $u_n>0$, the sequence $\left{s_n\right}$ is a monotone increasing sequence. As $\Sigma u_n$ converges to $s, \lim s_n=s$. Therefore the sequence $\left{s_n\right}$ is bounded above and $s_n \leq s$ for all $n \in \mathbb{N}$.

## 数学代写|实分析代写Real Analysis代考|Re-arrangement of terms

\begin{aligned} & \text { For example, let } f(n)=n+1 \text { if } n \text { be odd, } \ & =n-1 \text { if } n \text { be even. } \ & f(1)=2, f(2)=1, f(3)=4, f(4)=3, \cdots \cdots \ & \Sigma u_{f(n)}=u_2+u_1+u_4+u_3+\cdots \cdots \text { is a re-arrangement of } \Sigma u_n . \end{aligned}

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