Is there a function $f$ on $\mathbb{R}$ such that as $x \to 0$,


$$
f(x) = \sum\_{j=0}^N x^{1 - \frac{1}{j}} + o(x^{1- \frac{1}{N}}),
$$


for every $N \in \mathbb{N}$?
Heuristically there shouldn't be such a function, since the exponents accumulate at $1$ and thus higher terms are not "asymptotically independent", giving too much freedom in the choice of the coefficients in the sum, but I don't know how to even approach a proof.
It should be noted that if the exponents go to infinity, a modification of Borel's lemma holds yielding existence of such $f$.



Existence of function $f$ such that $f(x) \sim \sum_{j \in \mathbb{N}} x^{1 - \frac{1}{j}}$
First, $\sum\_{j=0}^N x^{1-1/j}$ is undefined, since $1/0$ is undefined. So, let us use $\sum\_{j=1}^N x^{1-1/j}$ instead.


For $x>0$, let
$$f(x):=\sum\_{j=1}^\infty x^{1-1/j}1(x<e^{-j})=\sum\_{1\le j<\ln(1/x)} x^{1-1/j}.$$
Then for each natural $N$ and small enough $x\in(0,1)$
$$0\le f(x)-\sum\_{j=1}^N x^{1-1/j}=\sum\_{N+1\le j<\ln(1/x)} x^{1-1/j}
\le x^{1-1/(N+1)}\ln(1/x)=o(x^{1-1/N})$$
as $x\downarrow0$, as desired.


