An upper bound of the form $A\cdot n\log n$ for the $n$th prime number

Question

It is well known that if $p_n$ is the $n$th prime number, then $$p_n>n\log n$$ and in fact, $p_n\sim n\log n$. I wanted to flip this inequality. That is, I wanted to find a real number $A$ such that $$p_n<A\cdot n\log n$$ From my calculations, $A$ should be greater than $1$, but less than $2$ (remember, as small as possible). I calculated the value of $A$ to be less than $1.33$. But, my computer is not much fast, so I couldn't find the smallest possible $A$. And note that $p_n<1.33n\log n$ doesn't work for all $n$, but it (probably) works for all $n>40000$. So my question is, is it true that $$p_n<1.33\,n\log n\quad\forall n>40000$$ and if yes, how can it be proved?
This inequality is indeed very good; since 1 is very near to 1.33. Any help would be appreciated.
Update: Weakening the bounds from this article gives $$p_n<1.37n\log n$$ I don't know if this would help, but a somewhat related procedure can probably prove this.

Answer 1

From $(1.7)$ in the paper cited by @JimmyK4542, it follows that $$ p_n < 1.1756\cdot n\log n $$ for all $n\geq 40000$. In fact, it also follows that $$ p_n < 1.2232\cdot n\log n $$ for all $n\geq 20$.