The sum $f(x)= \sum_{p \le x} e^{\frac{1}{p\log p}}$ is very close to $\pi(x)$. Why is that?

Question

The prime counting function $\pi(x)$ counts the number of primes less than a given $x$. There are other counting functions like Chebyshev's functions which count sums of logarithms of primes up to a given magnitude as well.

I am wondering why this summatory function (which is similar to Chebyshev's functions) over primes $p$ $$f(x)= \sum_{p \le x} e^{\frac{1}{p\log p}} $$ so closely approximates $\pi(x)$. For example, I wrote some python code that tells me $f(x)-\pi(x)<2$ for $x<10^7$.

Are there heuristics to support that $f(x)$ closely approximates $\pi(x)$?

I suspect $f(x)-\pi(x)$ grows slowly like $\log \log x$ or maybe even is bounded above by a constant. Thanks.

Answer 1

Since $e^{1/p\log p}=1+O(1/p\log p)$ and the sum

$$ \begin{aligned} \sum_p{1\over p\log p} &=\sum_{k\ge1}\sum_{2^k\le p<2^{k+1}}{1\over p\log p} \\ &\le\sum_{k\ge1}{1\over k2^k}\sum_{p\le 2^{k+1}}1=\sum_{k\ge1}{\pi(2^{k+1})\over k2^k} \\ &\le \text{const}\cdot\sum_{k\ge1}{2^{k+1}\over k2^k\log(2^{k+1})} \\ &\le \text{const}\cdot\sum_{k\ge1}{1\over k(k+1)}<\infty, \end{aligned} $$

we conclude that

$$ \sum_{p\le x}e^{1/p\log p}=\sum_{p\le x}1+O\left(\sum_{p\le x}{1\over p\log p}\right)=\pi(x)+O(1). $$