# Prime number inequality

I believe this is probably a high level undergraduate question, but i could easily be underestimating it and it's actually quite a bit higher than that.

I'm reading the Prime number theorem wikipedia page and i'm in part 4 under Proof sketch where sometime down they give in inequality:

x is a natural number, p is for prime's obviously: ##\sum_{x^{1-\epsilon }\geq p\geq x}^{} \log p \geq \sum_{p\leq x}^{} logx##

Where epsilon is any value greater than O. (O is some special value that they use in computer science a lot apparently, I might need to understand this value better to understand this inequality, i'm not sure.

Can somebody help me understand this inequality? C'mon I know there are some really smart people here!

Buzz Bloom
Gold Member
O is some special value that they use in computer science a lot apparently
Hi PsychonautQQ:

I am not into number theory, but I am almost certain that the "O" is zero, i.e., "0".

Regards,
Buzz

PsychonautQQ
fresh_42
Mentor
Hi PsychonautQQ:

I am not into number theory, but I am almost certain that the "O" is zero, i.e., "0".

Regards,
Buzz
My guess is ##O(something)##, e.g. ##O(1)\, , \,O(\log x) \, , \, O(\log \log x)## or even ##o(1)##. In any case I assume there is more than one typo in it, e.g. the ##\log x## in a summation over ##p## looks suspicious.

PsychonautQQ
mfb
Mentor
Here is a link to the article, the inequality discussed is the 6th formula in this section.
"for any ε > 0" is a comparison with the number 0. The big O notation (with the letter O) is mentioned because it is used later in the same line.

PsychonautQQ
Buzz Bloom
Gold Member
I found the equation in post #1 in the the linked article, although the post #1 version seems to have omitted a detail. It seems clear that the intent is
ε > zero​
and
O(x1-ε)​
is using the "big O" notation.

Regards,
Buzz

Last edited:
PsychonautQQ
mfb
Mentor
I thought it is the 6th formula there, but the right hand side is not identical.

PsychonautQQ
Wow, I am so smart for confusing 0 and O haha.

I still don't understand the inequality though, can somebody explain why it makes sense intuitively or link a proof to me?

mfb
Mentor
With the correct inequality from the Wikipedia article: We know ##x^{1-\epsilon} < p##. Take the logarithm on both sides and you get the inequality used in the article.

PsychonautQQ and Buzz Bloom
With the correct inequality from the Wikipedia article: We know ##x^{1-\epsilon} < p##. Take the logarithm on both sides and you get the inequality used in the article.
And we know that x^(1-epsilon) < p because of the domain which is being summed over? p is required to be bigger?

mfb
Mentor
We know it because the sum only runs over p satisfying this condition, right.