Very elementary proof of the prime number theorem 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.
 A: 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 $c=1$.
