Positivity of an alternating series. Greetings esteemed mathematicians.
I've managed to prove that the following series
\begin{equation}
f_{\lambda}(\omega)= \sum^{\infty}_{n=0}(-1)^n(2n+\lambda)\frac{\Gamma(n+\frac{1}{6})}{\Gamma(2n+1)}\omega^{2n}
\end{equation}
with $\lambda > 0$
converges $\forall \omega$. Does anyone have any suggestions as to how I can determine for which values of $\lambda$, $f_{\lambda}(\omega) > 0 \forall \omega$? Are there any known conditions for an alternating power series to converge to positive values?
Cheers,
Allen
 A: To start, let 
$$\hat{f}_{\lambda}(\omega) := \sum_{n \geq 0} (-1)^n (2n+\lambda) \frac{\omega^{2n}}{(2n)!} = \lambda \cos(\omega) - \omega \sin(\omega).$$
Now by an integral representation for the gamma function of the form 
$$\Gamma(n+1/6) = \int_0^{\infty} t^{n-5/6} e^{-t} dt,$$ if we integrate the first function termwise we get (after the substitution $s^2 = t$): 
$$f_{\lambda}(\omega) = \int_0^{\infty} 2 s^{1/3} \hat{f}_{\lambda}(\omega s) e^{-s^2} ds = \frac{1}{3} \Gamma \left(\frac{2}{3}\right) \left(3 \lambda \cdot \, _1F_1\left(\frac{2}{3};\frac{1}{2};-\frac{w^2}{4}\right)+2 w^2 \,
   _1F_1\left(\frac{5}{3};\frac{3}{2};-\frac{w^2}{4}\right)\right).$$ 
Now look at the graphs of the components of this last function corresponding to the coefficient of $\lambda$ and the remainder terms (plots shown below):
$\lambda$">
$\lambda$">
A little experimentation shows some examples which yield different signed behavior for $\omega$ around, and off, the origin:


There's probably a way to make this precise with calculus and letting $\lambda \equiv \lambda(\omega)$ depend on the other parameter, however, this shows the general method.
