Definitive proof about Sophie Germain primes

$ $When we write $\log$, we refer to natural logarithm. Let $i$, $k$, $n$ and $r$ be positive integers, $p $be a prime and $p_n$ be the $n$th prime. Initially, we prove that there are at least three odd integers $N$ of the form $2p+1$ that satisfy $2p_n< N<3p_n$ when $n≧9$ holds. Let $r$ be a positive integer and the number of $2p+1$ type odd integers in the range from $2p_n$ to $3p_n$. We consider the following inequality.
$$3p_n>2p_{n+2}+1\ …(1)$$
We assume that $3p_n≦2p_{n+2}+1$ holds. By Cramér's conjecture $p_{n+1}-p_n<(\log\ p_n)^2-\log\ p_n$ holds for $n≧5$,
$$p_n≦2(p_{n+2}-p_n)+1$$
$$p_n<2((\log\ p_{n+1})^2-\log\ p_{n+1}+(\log\ p_n)^2-\log\ p_n)+1$$
$$p_n<2((\log(p_n+(\log\ p_n)^2-\log\ p_n))^2-\log(p_n+(\log\ p_n)^2-\log\ p_n)+(\log\ p_n)^2-\log\ p_n)+1$$
holds. Let $f_1(x)$ be as follows.
$f_1(x)=2((\log(x+(\log\ x)^2-\log\ x))^2-\log(x+(\log\ x)^2-\log\ x)+(\log\ x)^2-\log\ x)+1-x$
$f_1(47)=0.79397596998018825465208164924441…$
$f_1(48)=0.35541303032635453018279531548847…$
$f_1(49)=-0.09160671627511924979677929262083…$
$f_1(50)=-0.54679890903296857822304045713304…$
$f_1(51)=-1.00989392734649122580129133427319…$
It is confirmed that $f_1(x)<0$ holds for $49≦x≦51$. $f_1(x)<0$ holds for $x≧49$ since the third term diverges faster than the first term. Therefore, the inequality (1) holds for $n≧16$ since the assumption is false. Let $f_2(n)=3p_n-2p_{n+2}-1$. The values of $f_2(n)$ from $1$ to $15$ are enumerated as follows.
$$\{-5,-6,-8,-6,-2,0,4,-2,6,12,10,24,28,22,22\}$$
From the above, it is proved that the inequality (1) holds for $n≧9$ and there are three or more odd integers of $2p+1$ type from $2p_n$ to $3p_n$ when $n≧9$ holds.
$ $We consider that $n≧9$ and $r≧3$ hold in this instance. Let $M$ be an odd integer and the product of the elements $2p_n+1$, $2p_{n+1}+1$, …, $2p_{n+r-1}+1$. We consider that the following equality holds for all $i$ such that $0≦i≦r-1$.
$$M=\prod_{j=0}^{r-1}(2p_{n+j}+1)$$
$$M≡\prod_{k=0, k≠i}^{r-1}(2p_{n+k}+1)\ (mod\ p_{n+i})$$
Let $a_i$ be a positive integer.
$$M=a_ip_{n+i}+\prod_{k=0,k≠i}^{r-1}(2p_{n+k}+1)$$
$$\prod_{j=0}^{r-1}(M-a_ip_{n+i})=M^{r-1}\ …(2)$$
$$M\prod_{k=0, k≠i}^{r-1}(M-a_k p_{n+k})≡M^{r-1}\ (mod\ p_{n+i})$$
$$(\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})-M^{r-2})M≡0\ (mod\ p_{n+i})$$
$ $We assume that $M≢0\ (mod\ p_{n+i})$ holds for all $i$ where $0≦i≦r-1$ holds after this. Let $b_i$ be a positive integer. We consider the following equation.
$$\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})=b_ip_{n+i}+M^{r-2}\ …(3)$$
By the equation (2),
$$(M-a_ip_{n+i})(b_ip_{n+i}+M^{r-2})=M^{r-1}$$
$$(M-a_ip_{n+i})b_i=a_iM^{r-2}$$
$$(b_i-a_iM^{r-3})M=a_ib_ip_{n+i}$$
holds. Let $c_i$ be a positive integer.
$$b_i=c_ip_{n+i}+a_iM^{r-3}\ …(4)$$
Comparing the equations (3) and (4), we find that the second term on the equation (4) is $a_i/ M$ times the one on the equation (3). If $M≢0\ (mod\ p_{n+i})$ holds for all $i$, then it becomes a contradiction since $0< a_i/ M<1$ holds and if this operation is repeated a finite number of times, then the terms will no longer be integers. Hence, the assumption is false. When $n≧9$ and $r≧3$ hold, $M≡0\ (mod\ p_{n+i})$ holds for at least one $i$.
$ $When $M≡0\ (mod\ p_{n+i})$ holds for some $i$, one of the elements in $M$ must be $p_{n+i}$ because $3p_{n+i}>3p_n$ holds and they cannot be three times or more odd multiples of $p_{n+i}$. From the above, it is proved that there are an infinite number of safe primes and Sophie Germain primes since at least one of the numbers $2p_n+1$, $2p_{n+1}+1$, …, $2p_{n+r-1}+1$ must be a safe prime when $n≧9$ and $r≧3$ hold and the sets of the numbers between $2p_n$ and $3p_n$ can be taken infinitely by increasing $n$. (Q.E.D.)