# Descartes' rule of signs for infinite series

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 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 $0<R<\infty$.

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).

$f$ 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 $$1 - \sum_{n=2}^\infty \frac{x^n}{n^2}$$
• So the condition may be restated as $\sum_{k=1}^{+\infty} a_k R^k>1$ (which is automatically true if $R=+\infty$) – Pietro Majer Sep 2 '18 at 7:21