Background
I had asked on a forum if my conjecture was of any value:
$$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f\left(\frac{rk}{n}\right)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_{0^+}^\infty f(x) \, dx $$
Where $f(x)$ is a smooth continuous function whose integral $\int_0^\infty f(x) d x $ is absolutely convergent. $d_r$ is the $r$-th number of an arbitrary sequence.
One of the commentor's said:
... For example for $d_r=\log(p)$ if $ r=p^k$ and $d_r=0$ if $r$ is not a prime power (the Von Mangoldt function) your formula is a famous theorem: the prime number theorem ....
He didn't give any further hints.
Question
Can someone elaborate on how this implies the PNT?
This has nothing to do with the PNT.
If $d_r \ge 0$ and $ \sum_{r \le y} d_r \sim ay$ and $f$ is continuous and $L^1$ then
$$\lim_{n \to \infty} \sum_{r=1}^n d_r f\left(\frac{rk}{n}\right)\frac{k}{n} =\lim_{n \to \infty} \sum_{r=1}^n a f\left(\frac{rk}{n}\right)\frac{k}{n} = a\int_0^k f(x)dx$$
On the other hand
$$\lim_{s \to 1^+} (s-1) \sum_{r=1}^\infty d_r r^{-s}=a$$
If $\sum_{r \le y} d_r = o(y)$ then both side converges to $0$, if $ y = o(\sum_{r \le y} d_r)$ then both side diverge to $\infty$.
If $\sum_{r \le y} d_r \sim ay$ you can use partial summation to remove the assumption $d_r \ge 0$