Property of alternating sign series

Question

I have an alternating series $\sum_{i=0}^{\infty}(-1)^{i+1}a_ix^i$, with $a_i\geq 0$ and the series is easy to check to converge for any $x>0$. I numerically checked that this sum is negative for any $x$ I tried, in fact it converges to $-\infty$ when $x$ converges to $+\infty$. Is there a systematic way to prove this property? Namely that $f(x)<0$ for any $x>0$.

I tried to use alternating series test, but did not succeed. It works fine with any fixed $x$, but not for all $x$'s.

An example would be $a_{2i+1}=\frac{1}{i!(i+1)!}$ and $a_{2i}=\frac{1}{i!i!}$, but I am interested in a systematic approach, not solving this particular case.

Answer 1

Let $x \geq 0$. Then $\frac{a_{2i}}{2}x^{2i}+\frac{a_{2i+2}}{2}x^{2i+2} \geq \sqrt{a_{2i}a_{2i+2}x^{2i}x^{2i+2}}=a_{2i+1}x^{2i+1}$.

Summing over all $i \geq 1$ yields $\sum_{k \geq 2}{(-1)^ka_kx^k} \geq a_2x^2/2$.

Thus, $$f(x)=-a_0/2-(a_0/2-a_1x+a_2x^2/2)-\left(\sum_{k \geq 2}{(-1)^ka_kx^k} - a_2x^2/2\right) \leq -(a_0/2-a_1x+a_2x^2/2).$$

The right-hand side is never positive and goes to $-\infty$.