Consider the function given by
$$f(x)=1-a\_1x-a\_2x^2-a\_3x^3-\cdots$$
where each $a\_k\geq0$ and some $a\_j>0$. If $f(x)$ is a polynomial then [Descartes' Rule of signs](https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs) tells us there is exactly one positive zero, i.e. root of $f(x)=0$. 


Assume $f(x)$ is a (real) power series with radius of convergence $0<R<\infty$.



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> **Question.** For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s). 
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Descartes' rule of signs for infinite series
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $\lim\_{x \to R-} f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - \sum\_{n=2}^\infty \frac{x^n}{n^2}$$


