### 数学代写|解析数论作业代写Analytic Number Theory代考|MATH4304

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## 数学代写|解析数论作业代写Analytic Number Theory代考|The fundamental theorem of arithmetic

Theorem 1.10 Fundamental theorem of arithmetic. Every integer $n>1$ can be represented as a product of prime factors in only one way, apart from the order of the factors.

Proof. We use induction on $n$. The theorem is true for $n=2$. Assume, then, that it is true for all integers greater than 1 and less than $n$. We shall prove it is also true for $n$. If $n$ is prime there is nothing more to prove. Assume, then, that $n$ is composite and that $n$ has two factorizations, say
$$n=p_{1} p_{2} \cdots p_{s}=q_{1} q_{2} \cdots q_{t} .$$
We wish to show that $s=t$ and that each $p$ equals some $q$. Since $p_{1}$ divides the product $q_{1} q_{2} \cdots q_{t}$ it must divide at least one factor. Relabel $q_{1}, q_{2}, \ldots, q_{t}$ so that $p_{1} \mid q_{1}$. Then $p_{1}=q_{1}$ since both $p_{1}$ and $q_{1}$ are primes. In (2) we may cancel $p_{1}$ on both sides to obtain
$$n / p_{1}=p_{2} \cdots p_{s}=q_{2} \cdots q_{t} .$$
If $s>1$ or $t>1$ then $1<n / p_{1}<n$. The induction hypothesis tells us that the two factorizations of $n / p_{1}$ must be identical, apart from the order of the factors. Therefore $s=t$ and the factorizations in (2) are also identical, apart from order. This completes the proof.

Note. In the factorization of an integer $n$, a particular prime $p$ may occur more than once. If the distinct prime factors of $n$ are $p_{1}, \ldots, p_{r}$ and if $p_{i}$ occurs as a factor $a_{i}$ times, we can write
$$n=p_{1}{ }^{a_{1}} \cdots p_{r}^{a_{r}}$$ or, more briefly,
$$n=\prod_{i=1}^{r} p_{i}^{a_{i}} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The series of reciprocals of the primes

Theorem 1.13 The infinite series $\sum_{n=1}^{\infty} 1 / p_{n}$ diverges.
Proof. The following short proof of this theorem is due to Clarkson [11]. We assume the series converges and obtain a contradiction. If the series converges there is an integer $k$ such that
$$\sum_{m=k+1}^{\infty} \frac{1}{p_{m}}<\frac{1}{2} .$$
Let $Q=p_{1} \cdots p_{k}$, and consider the numbers $1+n Q$ for $n=1,2, \ldots$ None of these is divisible by any of the primes $p_{1}, \ldots, p_{k}$. Therefore, all the prime factors of $1+n Q$ occur among the primes $p_{k+1}, p_{k+2} \ldots$ Therefore for each $r \geq 1$ we have
$$\sum_{n=1}^{r} \frac{1}{1+n Q} \leq \sum_{t=1}^{x}\left(\sum_{m=k+1}^{x} \frac{1}{p_{m}}\right)^{t},$$
since the sum on the right includes among its terms all the terms on the left. But the right-hand side of this inequality is dominated by the convergent geometric series
$$\sum_{t=1}^{x}\left(\frac{1}{2}\right)^{t} .$$
Therefore the series $\sum_{n=1}^{x} 1 /(1+n Q)$ has bounded partial sums and hence converges. But this is a contradiction because the integral test or the limit comparison test shows that this series diverges.

## 数学代写|解析数论作业代写Analytic Number Theory代考|The fundamental theorem of arithmetic

$$n=p_{1} p_{2} \cdots p_{s}=q_{1} q_{2} \cdots q_{t} .$$

$$n / p_{1}=p_{2} \cdots p_{s}=q_{2} \cdots q_{t} .$$

$$n=p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$$

$$n=\prod_{i=1}^{r} p_{i}^{a_{i}} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The series of reciprocals of the primes

$$\sum_{m=k+1}^{\infty} \frac{1}{p_{m}}<\frac{1}{2}$$

$$\sum_{n=1}^{r} \frac{1}{1+n Q} \leq \sum_{t=1}^{x}\left(\sum_{m=k+1}^{x} \frac{1}{p_{m}}\right)^{t}$$

$$\sum_{t=1}^{x}\left(\frac{1}{2}\right)^{t}$$

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