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An earlier question links to a paper of Erdos in which he says that it is "well-known" that the Prime Number Theorem is equivalent to

$(\prod_{p\leq n}p)^{1/n} \to e$ as $n\to \infty.$ **

Here is my confusion.

If $~\prod_{p\leq n}p \sim e^n$ or $e^{\log \prod p}= e^{\sum \log p} \sim e^n,$

(the last relation appears in the linked question, but I take responsibility for it) doesn't this imply that

(*) $\lim_{n \to \infty} (\sum_{p\leq n} \log p - n) = 0?$

Of course it's true that $\lim_{n \to \infty} \frac{\sum \log p}{n} =1 $ and I do not think that (*) is true. But I think we do have in general that

$$ e^{f(x)}\sim e^{g(x)} \implies \lim (f(x) - g(x)) = 0,$$ since $\lim \frac{e^f}{e^g}= e^{f-g} = 1 \implies \lim (f-g) = 0.$

Can someone tell me where I have goofed? Thanks!

**If someone could point me to a proof of this I would appreciate it --I don't see it in Apostol or Hardy & Wright).

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2 Answers 2

up vote 1 down vote accepted

$(\Pi_{ p \leq n} p)^{1/n} \rightarrow e$, as $n \rightarrow \infty$, doesn't imply $\Pi_{p \leq n} p \sim e^n$.

Example: $(2^n n)^{1/n} \rightarrow 2$, as $n\rightarrow \infty$, but $2^n n$ is not equivalent to $2^n$.

Indeed, $n^{1/n}=e^{\frac{\log n}{n}} \rightarrow 1$ because $\frac{\log n}{n} \rightarrow 0$ as $n \rightarrow \infty$

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I am pretty sure this is right and your counterexample is good. Then in the question I linked to, the OP seems to make this error too? (**) is from Erdos, but the following line appears in the other question. –  daniel Oct 4 '12 at 14:47
    
I don't know if (*) is true or not. –  francis-jamet Oct 4 '12 at 15:03
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The fact that it's equivalent is rather simple.

Recall $\pi(x)/(x/ln x) \rightarrow 1$, your limit is quivalent to:

$$\sum_{p\leq n} \log p / n \rightarrow 1$$

if you take $x=n \in \mathbb{N}$ then: $$\sum_{p\leq n} \log p /n \leq (\log n) \pi(n) /n \rightarrow 1$$

And to bound it from below, you notice that this fraction is greater than 1.

Not sure how to argue from below, I'll check next time.

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Thanks for fixing the typo--but I still don't know what question you are answering. The fact the "it's" equivalent...what is "it?" –  daniel Oct 4 '12 at 14:58
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