Proof that $\lim (\vartheta(x) - x) \neq 0 ?$

Question

Does

$$(1)\hspace{5mm}\lim_{x\to \infty} (~\sum_{p \text{ prime},~ p\leq x~} \log p - x) =0~ ?$$

I know that

$\vartheta(x) = \pi(x)\log(x) - \int_2^x \frac{\pi(t)}{t}dt$

A few calculations suggest (1) is not true and I know that $\vartheta(x) \sim x$ does not imply it.

Edit: This was motivated by Erdos' statement (A) that $(\prod p)^{1/n}\to e$ as $n \to \infty$ is equivalent to the PNT (here), which prompted someone to say (B) that $\prod p \sim e^n,$ which if true would imply (1). That $A \not\rightarrow B$ is shown by counterexample here and the answer here shows that B is false however we arrive at it.

Following the comments, I edited to make it clear that the sum is taken over the primes. If this is not right, please re-edit. — Nate Eldredge, Oct 05 '12 at 13:02
@daniel Just think about the graph of $\vartheta(x)$: it has a jump discontinuity at every prime of height $\log x$. It is therefore impossible for any smooth function to approximate $\vartheta(x)$ with better than $O(\log x)$ accuracy, and you are asking for $o(1)$. — Erick Wong, Oct 05 '12 at 13:54
@ErickWong: Yes. So I think Alfred Chern has expressed this well but I like a second opinion. I have been staring at several iterations of his proof and have lost a little objectivity. The proof was originally a lot longer and I wasn't sure how much he could omit, etc. — daniel, Oct 05 '12 at 14:07

Alfred Chern · Accepted Answer · 2012-10-05T11:19:10.457

1

$\lim_{x\to \infty}(~\sum_{p\leq x~}\log p-x)$ does not exist. We can prove it by contradiction.

If $\lim_{x\to \infty}(~\sum_{p\leq x~}\log p-x)$ exists, then

$$\begin{align*} 0=\lim_{x\to \infty}(~\sum_{p\leq(x+1)}\log p-(x+1))-\lim_{x\to \infty}(~\sum_{p\leq x}\log p-x)=\lim_{x\to \infty}~\sum_{x<p\leq(x+1)}\log p-1 \end{align*}$$

It means that $\lim_{x\to \infty}~\sum_{x<p\leq(x+1)}\log p=1$.

If there is a prime in $(x,x+1]$, then $\sum_{x<p\leq(x+1)}\log p>\log x$, if there is no prime in $(x,x+1]$, then $\sum_{x<p\leq(x+1)}\log p=0$, so $\lim_{x\to \infty}~\sum_{x<p\leq(x+1)}\log p$ cannot exist, it contradicts with the previous, so $\lim_{x\to \infty}(~\sum_{p\leq x~}\log p-x)$ cannot exist.

edited Oct 05 '12 at 11:19

answered Oct 04 '12 at 16:51

Alfred Chern

2,658

The sum given is supposed to be over all primes $p$. – lhf Oct 04 '12 at 16:54
@lhf: Are you sure? I think the sum should supposed to be all integers $p$. If supposed to be all primes $p$, the limit is also not exist and more easy to judge than before. – Alfred Chern Oct 04 '12 at 16:58
@AlfredChern, see http://en.wikipedia.org/wiki/Chebyshev_function. – lhf Oct 04 '12 at 17:01
@daniel: OK, then the problem will be much easier. – Alfred Chern Oct 04 '12 at 17:02
@lhf: I'm so sorry that I thought wrong and make a mistake. – Alfred Chern Oct 04 '12 at 17:05
@daniel: I'm so sorry that I thought wrong and make a mistake. – Alfred Chern Oct 04 '12 at 17:07
@daniel: I am using $[x]$ for integral part $x$, and ${x}$ for decimal. But my judgment is wrong. – Alfred Chern Oct 04 '12 at 17:13
@daniel: I know how to deal with it now and I'll typing it out later. – Alfred Chern Oct 05 '12 at 00:06
@daniel: Yes, you are right! But I think the rest also show another point of view. – Alfred Chern Oct 05 '12 at 11:20

Proof that $\lim (\vartheta(x) - x) \neq 0 ?$

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