# 数学分析|MATH0003 Analysis 1代写2023

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## Instructions:

We will start with the basic properties of real numbers and use them to prove the main results in elementary differential calculus. We assume familiarity with the properties of real numbers such as completeness, order, and the field axioms.

SEQUENCES:

A sequence is a function $f: \mathbb{N} \rightarrow \mathbb{R}$, where $\mathbb{N}$ is the set of natural numbers. We say that a sequence $(a_n)$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n – L| < \epsilon$ for all $n \geq N$. We write $\lim_{n \rightarrow \infty} a_n = L$.

a) Prove that if $c>1$, the sequence $a_n=\frac{c^n}{n !}$ is decreasing for $n \gg 1$.

a) To prove that the sequence $a_n=\frac{c^n}{n!}$ is decreasing for $n\gg1$, we need to show that $a_{n+1} < a_n$ for all $n\gg1$:

\begin{align*} a_{n+1} & = \frac{c^{n+1}}{(n+1)!} \ & = \frac{c}{n+1} \cdot \frac{c^n}{n!} \ & < \frac{c}{n+1} \cdot \frac{c^n}{n!} \quad \text{since } c>1 \ & = \frac{c}{n} \cdot \frac{c^n}{n!} \cdot \frac{n}{n+1} \ & < \frac{c}{n} \cdot \frac{c^n}{n!} \quad \text{since } \frac{n}{n+1} < 1 \ & = a_n \end{align*}

Therefore, $a_n$ is a decreasing sequence for $n\gg1$.

b) Prove $\lim _{n \rightarrow \infty} a_n=0$, by starting at some suitable point $N$ in the sequence and give an estimate of the size of the factors which allows you to use Theorem 3.4.)

b) To prove that $\lim_{n\rightarrow\infty}a_n=0$, we will use Theorem 3.4 which states that if there exist constants $C>0$ and $p>0$ such that $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq N$, where $N$ is some suitable point in the sequence, then $\sum_{n=1}^{\infty}a_n$ converges absolutely.

Let us first find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can take the ratio of consecutive terms:

$\frac{a_{n+1}}{a_n}=\frac{(n+1) !}{2^{n+1}(n+1) !}=\frac{1}{2^{n+1}}$

So we have $|a_{n+1}/a_n|\leq 1/2$, which means that we can take $C=1/2$. Also, we can take $p=1$ since $1/n$ is a decreasing function for $n\geq 1$. Thus, we have $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq 1$ with $C=1/2$ and $p=1$.

Now we need to find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can solve the inequality $1/2^n\leq \epsilon$, where $\epsilon>0$ is a small number that we want to be close to zero. Taking the logarithm of both sides, we get $n\ln(2)\leq \ln(1/\epsilon)$, which implies that $n\geq (\ln(1/\epsilon))/\ln(2)$. Therefore, we can take $N=\lceil(\ln(1/\epsilon))/\ln(2)\rceil$, where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$.

With these values of $C$, $p$, and $N$, we can use Theorem 3.4 to conclude that $\sum_{n=1}^{\infty}a_n$ converges absolutely, which implies that $\lim_{n\rightarrow\infty}a_n=0$.

c) Prove part (b) differently by considering the series $\sum_0^{\infty} a_n$.

c) Another way to prove that $\lim_{n\rightarrow\infty}a_n=0$ is to consider the series $\sum_{n=0}^{\infty}a_n$. We can use the ratio test to show that this series converges:

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim {n \rightarrow \infty} \frac{(n+1) !}{2^{n+1} n !}=\lim {n \rightarrow \infty} \frac{1}{2}=\frac{1}{2}<1$

Therefore, the series $\sum_{n=0}^{\infty}a_n$ converges, which implies that $\lim_{n\rightarrow\infty}a_n=0$ by the nth-term test for convergence.