4
$\begingroup$

Can someone tell me if anything is wrong with this proof? It seems too good to be true, as it was very easy to create.

$$ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =c $$ $$ \ln(x!)=\psi(x)+\psi(x/2)+\psi(x/3)+\psi(x/4)...\psi(x/x) $$ $$ \frac{\ln(x!)}{\psi(x)+\psi(x/2)+\psi(x/3)+\psi(x/4)...\psi(x/x)}=1$$ $$ \frac{\ln(x!)/x}{(\psi(x)+\psi(x/2)+\psi(x/3)+\psi(x/4)...\psi(x/x) )/x}=1$$ $$ \lim_{x \rightarrow \infty}\frac{\ln(x!)/x}{(\psi(x)+\psi(x/2)+\psi(x/3)+\psi(x/4)...\psi(x/x))/x}=1$$ $$ \lim_{x \rightarrow \infty}\frac{\ln(x!)/x}{(c+c/2+c/3+c/4+c/5...c/x)}=1$$ $$ \lim_{x \rightarrow \infty}\frac{\ln(x!)/x}{c \times \text{harmonic}(x)}=1$$ $$ \lim_{x \rightarrow \infty}\frac{\ln(x!)/x}{\text{harmonic}(x)}=1/c$$ $$ c=1$$

$\psi(x)$ is the first chebyshev function identity and can be found here.

  • 1
    It is useful to mention in your question the definitions and notations you use. For example, what is $\psi$? What is $harmonic$?2012-11-02

1 Answers 1

11

In the prime number theorem, the main difficulty is that it is hard to eliminate the case where the limit does not exist and instead oscillates between some values, i.e., a lot of effort is needed in justifying your first statement $$\displaystyle \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =c$$ You can get bounds for $A = \limsup_{x \to \infty} \dfrac{\psi(x)}{x}$ and $a = \liminf_{x \to \infty} \dfrac{\psi(x)}{x}$. In fact, Selberg proved that (before his elementary proof with Erdos) that $A+a = 2$. The crux in the case of PNT is showing that $A = a$.

Once you prove that this limit exists, then it is relatively easy to get its value to be $1$.

EDIT

A better way to write out what you have written would be as follows:

First prove that

\begin{align*} \sum_{d \leq N} \dfrac{\Lambda(d)}d &= \log N + \mathcal{O}(1) \end{align*} The proof for this goes as follows. We have that $\log(N!) = N \log N + \mathcal{O}(N)$. Also, \begin{align} \log(N!) & = \sum_{d \leq N} \Lambda(d) \left \lfloor \dfrac{N}d \right \rfloor = \sum_{d \leq N} \Lambda(d) \left( \dfrac{N}d + \mathcal{O}(1)\right)\\ & = N \sum_{d \leq N} \dfrac{\Lambda(d)}d + \mathcal{O} \left(\sum_{d \leq N} \Lambda(d)\right) = N \sum_{d \leq N} \dfrac{\Lambda(d)}d + \mathcal{O} \left(N\right) \end{align} Hence, $$\sum_{d \leq N} \dfrac{\Lambda(d)}d = \log N + \mathcal{O}(1)$$ Now use Abel summation technique or by writing $\sum_{d \leq N} \dfrac{\Lambda(d)}d$ as $\displaystyle \int_{2^-}^x \dfrac{\psi(t)}{t}dt$ and performing integration by parts to conclude that to conclude that if $\lim_{x \to \infty} \dfrac{\psi(x)}x = c$ exists, then $c=1$.

  • 2
    I remember reading somewhere that so and so proved if the limit exists then it is equal to one, why would someone even write that if this proof is so elementry? Also how do you log out on this site2012-11-02
  • 3
    Chebyshev proved that if the limit exists then it is 1.2012-11-03