The asymptotic density of a subset $A \subset\mathbb{N}$ is defined as $d(A) = \lim_{n\rightarrow\infty} \frac{A \cap \{1, 2, ... , n\}}{n}$.
Let $A_{i} = \{ k \in\mathbb{N} : k$ can be written as a sum of at most $i$ fifth powers of natural numbers $\}$.
When $i < 5$ I suspect that $d(A_{i}) = 0$. There is at most $n$ fifth powers $\leq n^{5}$. Therefore, we see that there are at most
${n \choose 1} + ... + { n \choose i} $ ~ $O(n^{i}$) many elements in $A_{i}$. For $i < 5$, this can be used to show $d(A_{i})$ = 0.
When $i = 5 $ it appears that the quantity $d(A_{5})$ can at most be $\frac{1}{5!}$.
How do I even dare to do this when $i = 6$? Can I put at least an upper bound to the density of $A_{6}$?
According to page 7 of https://www.personal.psu.edu/rcv4/Waring.pdf every sufficiently large number is conjectured to be expressible as a sum of 6 fifth powers, which would mean the natural density in the $i=6$ case is 1.
I don't know if further progress has been made since that survey was written in 2000, but it certainly means you shouldn't be able to upper bound it.