I accidentally collided with this "proof" of the prime number theorem that is so painfully short and simple I can't even believe it, and it yields the right result, and there doesn't seem to be anything wrong with it, but I'm also pretty sure that if it worked it would have been published already, so there must be something wrong with it, I'd like someone to tell me where it fails:
Let $\phi (x)$ equal to the sum of the reciprocals of primes less than or equal to x $\phi (x) = \sum_{n=2}^x \dfrac{a_n}{n}$ where a_n is 1 when $n$ is a prime and $0$ otherwise Using Abel's summation method, choosing $b_n$ as $a_n$ and $f(t)$ as $\dfrac{1}{t}$ , $A(x)$ becomes $π(x)$ and we get $$\phi (x) = \frac{π(x)}{x} + \int_2^x {\frac{π(t)}{t^2}}dt$$ We also know that $\phi (x)$ is asymptotically equal to $\log(\log x)$ which is equivalent to saying that $\phi (x) = \log(\log x) + o(\log(\log x))$ and this is equal to the expression given by Abel's method. $\frac{π(x)}{x}$ is $o(1)$, so when added to the other little $o$ you just have $o(\log(\log x))$ So we have $$\int_2^x {\dfrac{π(t)}{t^2}}dt = \log(\log x) + o\left(\log(\log x)\right)$$ So for this to be true, the integrant must be equal to $\dfrac{1}{t \log t} + o\left(\dfrac{1}{t \log t}\right)$ And this implies that $$π(t) = \dfrac{x}{\log x} + o\left(\dfrac{x}{\log x}\right)$$ which is the prime number theorem So what's failing here?