Consider
$$f(x) = \sum_{n=1}^{\infty} \frac{x^n}{p_n^{p_n}}$$
Where $p_n$ is the $n$ th prime.
What is a very good asympotic $g(x)$ for this function as $x$ goes to $+\infty$ ?
I want at least
$$\lim_{x \to +\infty} \frac{f(x)}{g(x)} = 1$$
I know for sufficiently large $x$ we have
$$\exp(\ln(x)^{1+v}) \ll f(x) \ll \exp(x^v)$$
for any fixed $v$ with $0<v$. (The upper bound is easy to show using Stirling's approximation for the factorial and the Maclaurin series for $\exp(x)$. The lower bound is also quite easy)
The estimate $\lim f(x)/g(x)=1$ might be too difficult to achieve given the known estimates of the $n$-th prime $p_n$. To see the results on the $n$-th prime, see this.
Let $\epsilon>0$ be given. We have $x^n/p_n^{p_n} = \exp(n\log x - p_n \log p_n)$.
Although precise asymptotic is out of reach, we may strengthen your lower and upper bounds using $$ n\log^2 n\leq \ p_n\log p_n\leq (1+\epsilon) n\log^2 n $$ for $n\geq N_{\epsilon}$ with effectively computable $N_{\epsilon}$.
For a lower bound, single term of the series when $n=\lceil \exp(\sqrt{(1-\epsilon)\log x})\rceil$ is by $\log n\sim \sqrt{(1-\epsilon)\log x}$ and $p_n\log p_n\leq (1+\epsilon)n\log^2 n$, we have $$ n\log x -p_n\log p_n \geq \log x \cdot \exp(\sqrt{(1-\epsilon)\log x}) $$ $$ -(1-\epsilon^2)\log x \exp(\sqrt{(1-\epsilon)\log x}) $$ $$ \geq \epsilon^2 \log x \cdot \exp(\sqrt{(1-\epsilon)\log x}) $$ Thus, we have for sufficiently large $x$, $$ \exp(\epsilon^2 \log x \cdot \exp(\sqrt{(1-\epsilon) \log x})\leq f(x). $$ For an upper bound, we use $n\log^2 n\leq p_n\log p_n$ and we split the sum into two cases $n>\exp(\sqrt{(1+\epsilon)\log x})$ and $n\leq \exp(\sqrt{(1+\epsilon)\log x})$.
The former case, we have $(1+\epsilon)\log x < \log^2 n$ yielding $$ \log x - \log^2 n < -\epsilon \log x. $$ Then the sum over $n$ in this range is bounded by $$ \sum_{n>\exp(\sqrt{(1+\epsilon)\log x})} x^{-\epsilon n}\leq 1/2 $$ using geometric series.
The latter case $n\leq \exp(\sqrt{(1+\epsilon)\log x})$, by applying upper bound of $e^{n\log x}$, the sum over this range is bounded by $$ \exp(\log x \cdot \exp(\sqrt{(1+\epsilon)\log x})\sum_{N_{\epsilon}\leq n\leq \exp(\sqrt{(1+\epsilon)\log x})} \frac1{\exp(n\log^2 n)} $$ The sum over $n$ is a convergent series. With suitable $N_{\epsilon}$, we can assume that the sum is $\leq 1/2$.
Therefore, we obtain an upper bound of $f(x)$ by combining the two sums: $$ f(x)\leq \exp(\log x \cdot \exp(\sqrt{(1+\epsilon)\log x}). $$
Putting these estimates together, we obtained $$ \exp(\epsilon^2 \log x \cdot \exp(\sqrt{(1-\epsilon) \log x}) $$ $$ \leq f(x) \leq \exp(\log x \cdot \exp(\sqrt{(1+\epsilon)\log x}). $$ The above estimates are in fact equivalent to a simpler expression. Given $\epsilon >0$, for sufficiently large $x$, $$\exp(\exp(\sqrt{(1-\epsilon)\log x}))\leq f(x)\leq \exp(\exp(\sqrt{(1+\epsilon)\log x})).$$