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1

Prove $\prod_p \frac{p^2}{p^2-1}=\frac{\pi^2}{6}$.

Proof: Consider the Euler product formula of the Riemann zeta function $\zeta(s)$:

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_p \left(1 - \frac{1}{p^s}\right)^{-1}$$

where the product traverses all prime numbers $p$, and this formula holds for $\text{Re}(s) > 1$.

Let $s=2$, we know $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. Substituting $s=2$ into the Euler product formula, we get:

$$\zeta(2) = \prod_p \left(1 - \frac{1}{p^2}\right)^{-1} = \frac{\pi^2}{6}$$

Now consider the infinite product given in the problem:

$$\begin{align} \prod_p \frac{p^2}{p^2-1} &= \prod_p \frac{1}{\frac{p^2-1}{p^2}} \\ &= \prod_p \frac{1}{1 - \frac{1}{p^2}} \\ &= \prod_p \left(1 - p^{-2}\right)^{-1} \end{align}$$

By comparison, this product is exactly the Euler product representation of $\zeta(2)$. Therefore,

$$\prod_p \frac{p^2}{p^2-1} = \zeta(2) = \frac{\pi^2}{6}$$

Q.E.D.

2

Prove that the series $\sum_p \frac{1}{p}$ diverges.

Proof: Use proof by contradiction. Assume that the series $\sum_p \frac{1}{p}$ converges. According to the Cauchy criterion, this means that for any $\epsilon > 0$, there exists $N$ such that for all $n > m \ge N$, $\sum_{k=m+1}^n \frac{1}{p_k} < \epsilon$. In particular, this means that the tail series $\sum_{k=K+1}^\infty \frac{1}{p_k}$ converges for any $K$ (as the limit of $n \to \infty$).

Let $K$ be any positive integer. Consider the set of positive integers $M_K = \{m \in \mathbb{N} \mid \forall p \text{ s.t. } p|m, p > p_K\}$ whose prime factors are all greater than $p_K$. Since $\sum_{k=K+1}^\infty \frac{1}{p_k}$ converges (by assumption), according to the relationship between infinite products and series, the infinite product $\prod_{k=K+1}^\infty (1 - \frac{1}{p_k})$ converges to a positive value $P_K' > 0$.

Therefore, the Euler product $\sum_{m \in M_K} \frac{1}{m} = \prod_{k=K+1}^\infty (1 - \frac{1}{p_k})^{-1} = \frac{1}{P_K'}$ converges to a finite value $V_K$.

Now, let $P_K = p_1 p_2 \cdots p_K$ be the product of the first $K$ prime numbers. Consider integers of the form $1 + q P_K$, where $q = 1, 2, 3, \dots$.

Any prime factor $p$ of $1 + q P_K$ must satisfy $p \nmid P_K$, otherwise $p \mid q P_K$ and $p \mid (1 + q P_K)$, which means $p \mid 1$, which is impossible.

Therefore, all prime factors of $1 + q P_K$ are greater than $p_K$, that is, $1 + q P_K \in M_K$ holds for all $q \ge 1$.

Thus, we have the series inequality:

$$\sum_{q=1}^\infty \frac{1}{1 + q P_K} \le \sum_{m \in M_K} \frac{1}{m} = V_K$$

This shows that if $\sum_p \frac{1}{p}$ converges, then the series $\sum_{q=1}^\infty \frac{1}{1 + q P_K}$ must converge.

However, we use the limit comparison test to compare the series $\sum_{q=1}^\infty \frac{1}{1 + q P_K}$ with the divergent harmonic series $\sum_{q=1}^\infty \frac{1}{q}$:

$$\lim_{q \to \infty} \frac{\frac{1}{1 + q P_K}}{\frac{1}{q}} = \lim_{q \to \infty} \frac{q}{1 + q P_K} = \frac{1}{P_K}$$

Since $P_K = p_1 \cdots p_K \ge 2$, the limit value $\frac{1}{P_K}$ is a positive finite constant.

Because the harmonic series $\sum_{q=1}^\infty \frac{1}{q}$ diverges, according to the limit comparison test, the series $\sum_{q=1}^\infty \frac{1}{1 + q P_K}$ must also diverge.

This contradicts the conclusion "$\sum_{q=1}^\infty \frac{1}{1 + q P_K}$ converges" derived from the assumption "$\sum_p \frac{1}{p}$ converges".

Therefore, the initial assumption "$\sum_p \frac{1}{p}$ converges" must be wrong. This means that the series $\sum_p \frac{1}{p}$ does not satisfy the Cauchy criterion, so the series diverges. Q.E.D.

3

Prove that the sequence $\{6n-1\}$ contains infinitely many prime numbers.

Proof: Use proof by contradiction. Assume that there are only finitely many prime numbers of the form $6n-1$, let them be $p_1, p_2, \ldots, p_r$. Consider the integer $N = 6(p_1 p_2 \cdots p_r) - 1$.

First, $N > 1$. In the prime factorization of $N$, all prime factors $p$ must satisfy $p \nmid 6$, that is, $p$ cannot be 2 or 3. Therefore, any prime factor $p$ of $N$ must be of the form $6k+1$ or $6k-1$.

Note that $N = 6(p_1 p_2 \cdots p_r) - 1 \equiv -1 \pmod 6$.

If all prime factors of $N$ are of the form $6k+1$, then their product $N$ must also be of the form $6k+1$. (Because $(6k_1+1)(6k_2+1) = 36k_1k_2 + 6k_1 + 6k_2 + 1 = 6(6k_1k_2+k_1+k_2)+1 \equiv 1 \pmod 6$) This contradicts $N \equiv -1 \pmod 6$.

Therefore, $N$ must have at least one prime factor of the form $6k-1$, let it be $p$.

We prove that $p$ is not equal to any of $p_1, p_2, \ldots, p_r$. If $p = p_i$ holds for some $i \in \{1, 2, \ldots, r\}$, then $p_i \mid N$ and $p_i \mid 6(p_1 p_2 \cdots p_r)$. Therefore $p_i$ must divide their difference, that is, $p_i \mid (6(p_1 p_2 \cdots p_r) - N)$, which is $p_i \mid 1$. This is impossible.

So, $p$ is a prime number of the form $6k-1$, but it is not in our assumed finite list $p_1, p_2, \ldots, p_r$. This contradicts our initial assumption (all prime numbers of the form $6n-1$ are in this list).

Therefore, the assumption is wrong, and there are infinitely many prime numbers of the form $6n-1$. Q.E.D.

5

Using $\prod_{p \leqslant x}\left(1-\frac{1}{p}\right)^{-1} \leqslant \prod_{K=2}^{\pi(x)+1}\left(1-\frac{1}{K}\right)^{-1}$, prove: (1) $\pi(x)>\log x-1$; (2) $p_n<3^{n+1}$ ($p_n$ is the $n$-th prime number).

Proof:

(1) Prove $\pi(x) > \log x - 1$

We know that for $x \ge 1$, $\sum_{n \le x} \frac{1}{n} > \log x$. At the same time, we have

$$\sum_{n \le x} \frac{1}{n} \le \sum_{n \in S_x} \frac{1}{n} = \prod_{p \le x} \left(1 - \frac{1}{p}\right)^{-1}$$

where $S_x$ is the set of positive integers whose prime factors are all $\le x$.

Combining the above inequality and the inequality given in the problem, we get:

$$\log x < \sum_{n \le x} \frac{1}{n} \le \prod_{p \le x} \left(1 - \frac{1}{p}\right)^{-1} \le \prod_{K=2}^{\pi(x)+1} \left(1 - \frac{1}{K}\right)^{-1}$$

Calculate the product on the right:

$$\begin{align} \prod_{K=2}^{\pi(x)+1} \left(1 - \frac{1}{K}\right)^{-1} &= \prod_{K=2}^{\pi(x)+1} \left(\frac{K-1}{K}\right)^{-1} \\ &= \prod_{K=2}^{\pi(x)+1} \frac{K}{K-1} \\ &= \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{\pi(x)+1}{\pi(x)} \\ &= \pi(x) + 1 \end{align}$$

Therefore, we get $\log x < \pi(x) + 1$. Rearranging gives $\pi(x) > \log x - 1$.

(2) Prove $p_n < 3^{n+1}$

From (1) we know $\pi(x) > \log x - 1$. Let $x = p_n$, where $p_n$ is the $n$-th prime number. Then $\pi(x) = \pi(p_n) = n$. Substituting into the inequality, we get:

$$n > \log p_n - 1$$

Rearranging gives:

$$\log p_n < n + 1$$

Taking the exponent on both sides (with natural logarithm base $e$):

$$p_n < e^{n+1}$$

Since $e \approx 2.718 < 3$, we have $e^{n+1} < 3^{n+1}$. Therefore,

$$p_n < 3^{n+1}$$

Q.E.D.

References

• https://zhuanlan.zhihu.com/p/543010816

• https://zhuanlan.zhihu.com/p/544792225

• https://zhuanlan.zhihu.com/p/545595129

• https://zhuanlan.zhihu.com/p/546385318

• https://zhuanlan.zhihu.com/p/547333984

• https://zhuanlan.zhihu.com/p/548061254

Contributors: Ming Wei

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