Here's a proof of the positivity of $$ c_n(\alpha) := \sum_{r=0}^n (-1)^r {n\choose r}^\alpha $$ for all even $n$ and real $\alpha < 1$. It follows (via M.Wildon's clever $F(x) F(-x)$ trick at mo.84958) that $\sum_{n=0}^\infty x^n / n!^{\alpha} > 0$ for all $x \in\bf R$.
The key is to write $c_n(\alpha)$ as a finite difference $$ \sum_{r=0}^n (-1)^r {n\choose r} \cdot {n\choose r}^{\alpha - 1} $$ and show that the Gamma interpolation $$ \bigl(\Gamma(r+1)\Gamma(n-r+1) / n!\bigr)^{1-\alpha} = n!^{\alpha-1} \exp\bigl((1-\alpha) (\log\Gamma(r+1) + \Gamma(n-r+1)\bigr) $$ of ${n\choose r}^{\alpha - 1}$ has a positive $n$-th derivative for all $r \in [0,n]$.
This in turn follows from the fact that the expansion of $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ in a Taylor series about $r = n/2$ has positive $(r - (n/2))^k$ coefficient for each $k=2,4,6,\ldots$. [The coefficient vanishes for odd $k$ because $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is an even function of $r-(n/2)$.] Indeed the well-known formula $$ \log \Gamma(x) = -\gamma x - \log x + \sum_{k=1}^\infty \left[ \frac{x}{k} - \log \left( 1 + \frac{x}{k} \right) \right] $$ shows that the $k$-th derivative of $\log\Gamma(x)$ is positive for all $x>0$ and $k=2,4,6,\ldots$, because this is true for $-\gamma x - \log x$ and for each term in the sum. Therefore in the Taylor expansion $$ \log \Gamma(r+1) = \log(n/2)! + \sum_{k=1}^\infty \phantom. g_k (r-(n/2))^k $$ each of $g_2,g_4,g_6,\ldots$ is even. Since $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is $$ 2\log(n/2)! + 2 \Bigl( g_2 (r-(n/2))^2 + g_4 (r-(n/2))^4 + g_6 (r-(n/2))^6 + \cdots\Bigr), $$ the claim follows. Multiplying by $1 - \alpha$ and substituting into the exponential series, we deduce that $(\Gamma(r+1) \Gamma(n-r+1))^{1-\alpha}$, too, is a positive combination of even powers of $r-(n/2)$.
Now if a function $g$ has positive $n$-th derivative, then its first finite difference $$ g(x+1) - g(x) = \int_x^{x+1} g'(y) dy $$ has positive $(n-1)$-st derivative; repeating this argument $n$ times, we find that the $n$-th finite difference is positive, and we're done.