# Bound in Prime Numbers

I am currently working on Theorem 1.3 below. But, I was confused why $T_2$ is less than the product of primes( referring to the inequality that is encircle). It was stated in the proof that if $p$ satisfies $3n<p \leq n$ and also if $p$ satisfies $\frac{3n}{2}<p \leq 2n$ , it's power is $0$. I think that $T_2$ is subdivide into intervals but I don't get why it is less than. In my own understanding, it is equal. But, I guess I was wrong. Can someone help me out? This is the reference (http://www.m-hikari.com/ijcms-password/ijcms-password13-16-2006/elbachraouiIJCMS13-16-2006.pdf)

• This is fuzzy enough it is hard to read – Ross Millikan Aug 24 '17 at 17:10
• @RossMillikan I edited, copied the url for the jpeg, cancelled edit, pasted url into browser, downloaded that and finally printed. Clear enough; I was curious. It would be nice to know the source of this excerpt and what reference [3] is. – Will Jagy Aug 24 '17 at 17:13
• Sorry for that sir. But you can view the picture alone. It is clearer. – jENEVA Aug 24 '17 at 17:14