Let $f: \mathbb{N} \longrightarrow [0, \infty)$ be a decreasing function. So $$\sum_{p\ prime}f(p) $$ converges if, and only if $$\sum_{n=2}^{\infty}\dfrac{f(n)}{\log_e(n)}$$ converges.
I should use the Chebyshev's Theorem $$(\log_e(2) - o(1))\dfrac{x}{\log_e(x)} < \pi(x) < (2\log_e(2) + o(1))\dfrac{x}{log_e(x)}$$ but i don't know how.
Maybe i can use this corollary $$(1 - o(1))\dfrac{nlog_e(n)}{2\log_e(2)} < p_n < (1 + o(1))\dfrac{n\log_e(n)}{\log_e(2)} $$ where $p_n$ is the n-th prime.
Let $\pi (x)$ be the number of primes less than $x.$ Let $P(x)$ be the set of primes less than $x.$
Extend the domain of $f$ to $[1,\infty)$ such that $f$ is continuous and decreasing.
It is easily shown that the Prime Number Theorem $1=\lim_{x\to \infty}\frac {\pi (x)\ln x}{x}$ is equivalent to $\pi(x)\sim li(x)=\int_2^x \frac {1}{\ln t}dt\,,$ that is, $\lim_{x\to\infty}\pi (x)/li(x)=1.$ And then it is easy to show that if $f:[2,\infty)\to\Bbb R $ is continuous & monotonic and if $f(x)=O(x^k)$ as $x\to \infty$ for some $k>0$ then $\sum_{p\in P(x)}f(p)\sim \int_2^x\frac {f(t)}{\ln t} dt.$
Now with $g(x)=\frac {f(x)}{\ln x},$ if $f:[2,\infty)\to [0,\infty)$ is continuous & decreasing then $g :[2,\infty)\to [0,\infty)$ is continuous & decreasing, and an elementary calculus result (Integral Test) is that $\sum_{n=2}^{\infty}g(n)$ converges iff $\int_2^{\infty}g(t)dt$ converges.