Is it possible to get the closed form of the below series? $$\sum_{k=0}^{\infty}(-1)^k\frac{x^{p(k+1)}}{\Gamma(p(k+1)+1)}$$ with $p \in (0, 1)$, and $\Gamma(x)$ is the Gamma function.
I know that without $p$, the formula is just $1-e^{-x}$.
And I also tried substitution $y = x^p$.
In addition, searching stackexchange and google, I think this may relate to generalized binomial theorem.
$$f_p(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{p(k+1)}}{\Gamma[p(k+1)+1]}$$
The very first ones are "simple" $$f_1(x)=1-e^{-x} \qquad f_2(x)=1-\cos (x)$$ $$f_3(x)=-\frac{1}{3} e^{-x} \left(1-3 e^x+2 e^{3 x/2} \cos \left(\frac{\sqrt{3} x}{2}\right)\right)$$ $$f_4(x)=1-\cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right)$$ For $p>4$, the only expressions are generalized hypergeometric functions. I give you the very first in order you see the patterns $$f_5(x)=1-\, _0F_4\left(;\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};-\left(\frac{x}{5}\right)^5 \right)$$ $$f_6(x)=1-\, _0F_5\left(;\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};-\left(\frac{x}{6}\right)^6\right)$$ $$f_7(x)=1-\, _0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac {6}{7};-\left(\frac{x}{7}\right)^7\right)$$ making the general form to be $$f_p(x)=1-\, _0F_{p-1}\left(;\frac{1}{p},\frac{2}{p},\cdots,\frac{p-1}{p};-\left(\frac{x}{p}\right)^p\right)$$