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I am interested in finding out whether we can tell if the function is positive or negative based on its power series coefficients.

Specifically, I am interested in the case of $x>0$ and \begin{align} f(x)=\sum_{n=1}^\infty a_n x^n = a_1x+a_2x^2+... \end{align} where $a_n$'s have an alternating signs and $a_1>0$. Also, assume that the series is convergent for all $x \ge 0$.

It is easy to see that $f(x)>0$ is positive in the neighborhood of zero for $x>0$.

So, my question can we say more based on the coefficients?

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  • $\begingroup$ ...It should be negative... Look at the behavior around $x=0$... $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 17:17
  • $\begingroup$ @SimplyBeautifulArt Why negative around zero? we assume that $a_1>0$. $\endgroup$ – Boby Jan 25 '17 at 18:51
  • $\begingroup$ If $a_1>0$ and the series converges in a neighborhood of zero, then I think $f(x)$ has the same sign as $x$ in a neighborhood of zero. The alternating condition seems redundant. On the other hand the given conditions seem insufficient to guarantee $f(x)>0$ for all $x>0.$ $\endgroup$ – David K Jan 25 '17 at 21:34
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Notice $\lim\limits_{x\to 0} \frac{f(x)}{x}=a_1$. So $f(x)$ must be negative for negative values close to $0$,

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  • $\begingroup$ Sure. The behavior around zero can be predicted. The question is whether we can say more? $\endgroup$ – Boby Jan 25 '17 at 18:55
  • $\begingroup$ what do you mean, no function of the form you are interested in is positive or negative. $\endgroup$ – Jorge Fernández Hidalgo Jan 25 '17 at 18:56
  • $\begingroup$ Why not? Take $a_1=1, a_2=-2, a_2=5$ and $a_n=0$ for $n>3$. Surely this function is positive for $x>0$. I guess the confusion is that I want to show it only for positive $x$. $\endgroup$ – Boby Jan 25 '17 at 18:59
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If $|a_n|\cdot x$ approaches $0$ monotonically for $x\in[0,r]$, then it follows that for $x\in[0,r]$,

$$a_1x-a_2x^2<\sum_{n=1}^\infty a_nx^n<a_1x$$

which follows from the alternating test.

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