Can we use $n\log n$ instead of $n$-th prime?

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Denote $\pi(x)$ be the number of primes $\leq x,$ $p(n)$ be the $n$-th prime number. We have $\pi(p(n))=n.$

It's well known that $$\pi(x)\sim \frac{x}{\log x} \\p(n)\sim n\log n.$$

Is it always true that if $f(x)$ is a function and

  • $f(x)>0, \forall x>0$

  • $f(x)$ is a monotonic function

  • $f(x)=O(x^r)$ for some $r>0$

  • $\sum_{n=1}^{\infty}f(n)=\infty$

then $$\sum_{p\leq x}f(p)\sim \sum_{t\leq \pi(x)}f(t\log t)$$?