As the title ask, what is the best lower and upper bounds for the product below : $$ \prod \limits_{x < p \leq y} \frac{p+1}{p}$$ such that $p$ denote the prime numbers in which fullfil the conditions under the product ?

2$\begingroup$ Why do you not simply ask for upper and lower bounds for (1), since (2) is the reciprocal of (1)? Also, it's best to use a more descriptive title. You can use LaTeX in the title, so you could change it to "Best known upper and lower bounds for $\prod_{x<p\le y} (p+1)/p$. $\endgroup$– Joe SilvermanNov 29, 2016 at 19:35

$\begingroup$ you are right ! $\endgroup$– AhmadNov 29, 2016 at 19:41
3 Answers
Your product is the following $$\prod_{y<p\leq x} \left(1+\frac{1}{p}\right).$$ You can use the fact that $$\prod_{y<p\leq x} \left(1+\frac{1}{p}\right)\leq \prod_{y<p\leq x} \left(1\frac{1}{p}\right)^{1} \leq \frac{\pi^2}{6} \prod_{y<p\leq x} \left(1+\frac{1}{p}\right).$$ Note that $\frac{\pi^2}{6}=\frac{1}{\zeta(2)},$ where $\zeta$ is the Riemann zeta function. According to the best current explicit results of Pierre Dusart: For $x > 1$, $$\prod_{p\leq x} \left(1\frac{1}{p}\right)< \frac{\exp(\gamma)}{\ln{x}} \left(1+\frac{0.2}{\ln^3{x}}\right)$$ and for $x \geq 2278382,$ $$\prod_{p\leq x} \left(1\frac{1}{p}\right)> \frac{\exp(\gamma)}{\ln{x}} \left(1\frac{0.2}{\ln^3{x}}\right)$$ where $\gamma$ is the Euler's constant $(\gamma \sim 0.5772157)$ and you are done!

$\begingroup$ In Dusart's most recent paper (which I cited in my answer), one gets $\ln(x)^3$ instead of $\ln(x)^2$ in the denominator of the error term (at the cost of changing $x\geq 2973$ to $x\geq 2278382$). I'm only pointing this out so that those who come to MathOverflow and see this is the accepted answer don't think this is the best bound available. $\endgroup$ Nov 29, 2016 at 20:39

$\begingroup$ @Pace Nielsen oh yes you are right! I will edit the post. $\endgroup$ Nov 29, 2016 at 20:41

$\begingroup$ Also, you don't want to use Dusart's bounds for $\prod_{p\leq x}\left(1\frac{1}{p}\right)$, but rather his bounds for the reciprocal (which are not the same as reciprocating the bounds of the reciprocal). $\endgroup$ Nov 29, 2016 at 20:42

3$\begingroup$ In the second inequality, the factor $\frac{\pi^2}6$ is quite wasteful for nontrivial $y$; a better bound for $\prod_{p>y} \big( 1\frac1{p^2} \big)^{1}$ is $\frac y{y1}$, resulting from the telescoping product $\prod_{n>y} \big( 1\frac1{n^2} \big)^{1} = \frac{\lfloor y\rfloor +1}{\lfloor y\rfloor}$. $\endgroup$ Nov 29, 2016 at 23:31
The best source for explicit, unconditional bounds on such products that I'm aware of is in the work of Pierre Dusart. See his paper Explicit estimates of some functions over primes.
In Section 5.4, Theorem 5.9, he gives upper and lower bounds for $$ Q(x):=\prod_{p\leq x}\frac{p}{p1}. $$ Using these, we can get bounds on $$ Q(y)/Q(x)=\prod_{x< p\leq y}\frac{p}{p1}. $$ Notice that the fraction you want is $$ \frac{p+1}{p}=\frac{p}{p1}\cdot \left(1\frac{1}{p^2}\right) $$ so you have reduced to bounding $$ \prod_{x<p\leq y}\left(1\frac{1}{p^2}\right) $$ which can be done by standard techniques.
(Note that Dusart's bounds are for $x\geq 2278382$, but for small values of $x$ and $y$ you can do an explicit computation.) Alternatively, you could work directly with the fraction you want (instead of $p/(p1)$) and use the same techniques as Dusart.
Though Dusart bounds are good ones for $\,\prod\limits_{p \le x}{\left(1 1/p\right)}$, I think a better idea to the prime product asked by Ahmad is to explore the Dedekind psi function, used in the theory of modular functions, defined as $\,\psi(n)/n = \prod\limits_{pn}\left( 1 +\frac{1}{p} \right)$.