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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.

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    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.2012-10-05
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    @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 $o(1)$.2012-10-05
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    @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.2012-10-05

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

It means that $\lim_{x\to \infty}~\sum_{x.

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

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    The sum given is supposed to be over all *primes* $p$.2012-10-04
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    @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.2012-10-04
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    @AlfredChern, see http://en.wikipedia.org/wiki/Chebyshev_function.2012-10-04
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    @daniel: OK, then the problem will be much easier.2012-10-04
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    @lhf: I'm so sorry that I thought wrong and make a mistake.2012-10-04
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    @daniel: I'm so sorry that I thought wrong and make a mistake.2012-10-04
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    @daniel: I am using $[x]$ for integral part $x$, and $\{x\}$ for decimal. But my judgment is wrong.2012-10-04
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    @daniel: I know how to deal with it now and I'll typing it out later.2012-10-05
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    @daniel: Yes, you are right! But I think the rest also show another point of view.2012-10-05