2
$\begingroup$

Does this inequality always hold : $$\frac{1}{6} \pi ^2 \prod _{i=1}^x \frac{\left(p_i\right){}^2-1}{\left(p_i\right){}^2}\leq \frac{1}{p_x}+1 $$

such that $p_i$ is the $i$-th prime number

$\endgroup$
  • 5
    $\begingroup$ Two suggestions. First, you should add the number-theory tag to questions of this sort. Second, you seem to be asking a lot of questions of a similar nature in a short amount of time. You got a complete answer to one of your questions, and it seems to me that the methods used there would similarly be useful for this question. So until you spend a week or two working on it yourself, you probably shouldn't jump right in and ask it on MathOverflow. $\endgroup$ – Joe Silverman Sep 14 '16 at 22:14
6
$\begingroup$

Yes. We have $$ \frac{1}{6} \pi ^2 \prod _{i=1}^x \frac{\left(p_i\right){}^2-1}{\left(p_i\right){}^2}=\prod_{i>x} \frac{p_i^2}{p_i^2-1}\leqslant \prod_{n=p_x+1}^{\infty} \frac{n^2}{n^2-1}=\frac1{p_x}+1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.