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# 数学代写|实分析代写Real Analysis代考|Monotone sequence

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## 数学代写|实分析代写Real Analysis代考|Monotone sequence

A real sequence ${f(n)}$ is said to be a monotone increasing sequence if $f(n+1) \geq f(n)$ for all $n \in \mathbb{N}$.

A real sequence ${f(n)}$ is said to be a monotone decreasing sequence if $f(n+1) \leq f(n)$ for all $n \in \mathbb{N}$.

A real sequence ${f(n)}$ is said to be a monotone sequence if it is either a monotone increasing sequence or a monotone decreasing sequence.
Note. If $f(n+1)>f(n)$ for all $n \in \mathbb{N}$, the sequence ${f(n)}$ is said to be a strictly monotone increasing sequence.

If $f(n+1)<f(n)$ for all $n \in \mathbb{N}$, the sequence ${f(n)}$ is said to be a strictly monotone decreasing sequence.

If for some natural number $m, f(n+1) \geq f(n)$ for all $n \geq m$ the sequence ${f(n)}$ is said to be an ‘ultimately’ monotone increasing sequence.

If for some natural number $m, f(n+1) \leq f(n)$ for all $n \geq m$ the sequence ${f(n)}$ is said to be an ‘ultimately’ monotone decreasing sequence.

## 数学代写|实分析代写Real Analysis代考|Some important sequences

1. The sequence $\left{\left(1+\frac{1}{n}\right)^n\right}$ is a monotone increasing sequence, bounded above.
Let $u_n=\left(1+\frac{1}{n}\right)^n$. Then $u_{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}$.
Let us consider $n+1$ positive numbers $1+\frac{1}{n}, 1+\frac{1}{n}, \cdots, 1+\frac{1}{n}(n$ times) and 1.
Applying A.M.> G.M., we have $\frac{n\left(1+\frac{1}{n}\right)+1}{n+1}>\left(1+\frac{1}{n}\right)^{\frac{n}{n+1}}$
\begin{aligned} & \text { or, }\left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1}{n}\right)^n \text {. } \ & \text { i.e., } u_{n+1}>u_n \text { for all } n \in \mathbb{N} \text {. } \end{aligned}
This shows that the sequence $\left{u_n\right}$ is a monotone increasing sequence.
\text { Now } \begin{aligned} u_n & =1+1+\frac{n(n-1)}{2 !} \frac{1}{n^2}+\cdots+\frac{n(n-1) \cdots 2 \cdot 1}{n !} \frac{1}{n^n} \ & =1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\cdots+\frac{1}{n !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots \frac{2}{n} \cdot \frac{1}{n} \ & <1+1+\frac{1}{2 !}+\cdots+\frac{1}{n !} \text { for all } n \geq 2 . \end{aligned} We have $n !>2^{n-1}$ for all $n>2$. Utilising this
$$1+1+\frac{1}{2 !}+\cdots+\frac{1}{n !}<1+1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}} \text {, for } n>2 \text {. }$$
Also $1+1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}}=1+2\left[1-\left(\frac{1}{2}\right)^n\right]<3$ for all $n \in \mathbb{N}$. It follows that $u_n<3$ for all $n \in \mathbb{N}$, proving that the sequence $\left{u_n\right}$ is bounded above.

Thus the sequence $\left{u_n\right}$ being a monotone increasing sequence bounded above, is convergent. The limit of the sequence is denoted by $e$.
Since $u_1=2$, it follows that $2<u_n<3$ for all $n \geq 2$.

## 数学代写|实分析代写Real Analysis代考|Some important sequences

\begin{aligned} & \text { or, }\left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1}{n}\right)^n \text {. } \ & \text { i.e., } u_{n+1}>u_n \text { for all } n \in \mathbb{N} \text {. } \end{aligned}

\text { Now } \begin{aligned} u_n & =1+1+\frac{n(n-1)}{2 !} \frac{1}{n^2}+\cdots+\frac{n(n-1) \cdots 2 \cdot 1}{n !} \frac{1}{n^n} \ & =1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\cdots+\frac{1}{n !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots \frac{2}{n} \cdot \frac{1}{n} \ & <1+1+\frac{1}{2 !}+\cdots+\frac{1}{n !} \text { for all } n \geq 2 . \end{aligned}我们有$n !>2^{n-1}$表示所有的$n>2$。利用这个
$$1+1+\frac{1}{2 !}+\cdots+\frac{1}{n !}<1+1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-1}} \text {, for } n>2 \text {. }$$

## MATLAB代写

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