Consider the function $f(\cdot)$ defined as follows, $$ f(x) = \sum_{k=0}^{\infty} a_k \left(\frac{-1}{x}\right)^k$$ where $a_0 = 1$ and $a_k > 0$ for all $i$. Assume the series converges in absolute scence for all values of $x\in \mathbb{C}\setminus \{0\}$. Moreover, it has no complex root (and obviously, no negative root). I want to show it has a positive root. (I know it is not correct in general, example: $exp(-x^{-1})$) The coefficients are given as follows: $$ a_k := a_k(0)\\ a_0(y) = 1, \qquad \forall y\in\mathbb{R}_+\\ a_k(y) = \int_{y}^\infty (z-y) a_{k-1}(z) d\mu(z)\qquad\forall k>0 $$ where the measure $\mu$ is a finite measure on $\mathbb{R}_+$, and its Radon-Nikodym derivative wrt Lebesgeue measure is bounded by $C\times \exp(-az)$ (BTW, is there any especial name for such a measure?). Any suggestion is really helpful.
Here is what I have: It is evident that $lim_{x\to 0^{-}} f(x) = \infty$. Moreover, because of the choice of $a_k$s, if $f(\cdot)$ has no positive root, then the followings are correct:
$$ \lim_{x\to 0^{+}} f(x) = 0\\ f(x)> 0\qquad\forall x>0 $$
Finally, it is easy to prove: $$a_k\leq \frac{C^k}{k!k!}$$