# converge inequality for squares of primes

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

• 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. – Joe Silverman Sep 14 '16 at 22:14

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