Evaluate the sum
$$\sum_{k=0}^\infty\frac{(-1)^kz^{ak}}{\Gamma(1+ak)}$$
where $a\in \mathbb{R}_{>0}$ and $z\in\mathbb{C}$.
I know if $a=2$ then this is the series expansion for $\cos(z)$. But for arbitrary positive real $a$ I have no idea what this is.
This is a Mittag-Leffler function : $$E_{\alpha, \beta} (z) = \sum_{k=0}^\infty \frac{z^k}{\Gamma(\alpha k + \beta)},\quad \alpha,\beta\in\mathbb{C},\;\Re(\alpha)>0,\,\Re(\beta)>0, z\in \mathbb{C}$$ or in your case ($\beta=1$ is implicit) : $$E_{\,a} (-z^a) = \sum_{k=0}^\infty \frac{(-z^a)^k}{\Gamma(1+a k)}$$
with a table for the first values of $a$ : \begin{array} {c|c} a&E_a(x)\\ \hline 0&\frac 1{1-x}\\ 1&e^x\\ 2&\cosh\sqrt{x}\\ 3&\frac 13\left(e^{x^{1/3}}+2\,e^{-\frac 12 x^{1/3}}\cos\left(\frac{\sqrt{3}}2x^{\frac 13}\right)\right)\\ 4&\frac 12\left(\cos(x^{1/4})+\cosh(x^{1/4})\right)\\ \end{array}
Of course the corresponding expressions for $E_a(z^a)$ for $a$ a positive integer would be somewhat simpler and this for a good reason : we are simply using the Taylor series of the exponential $\;\displaystyle \exp(z):=\sum_{k=0}^\infty \frac {z^k}{k!}\;$ and keeping only the terms with $k$ multiple of $\,a\,$ which may be rewritten as :
$$E_a(z^a)=\frac 1a\sum_{k=0}^{a-1}\exp\left(z\;e^{\dfrac{2\pi i k}a}\right)$$
(for more details see this answer )
For $a=\frac n2$ you should get the hypergeometric function proposed by Claude for $x=-z^a$ $$E_{\frac n2}(x)= _0F_{n-1}\left(;\frac{1}{n},\frac{2}{n},\cdots,\frac{n-1}n;\frac{x^2}{n^n}\right)+\frac {2^{(n+1)/2}}{n!\sqrt{\pi}} {}_1F_{2n-1}\left(1;\frac{n+2}{2n},\frac{n+3}{2n},\cdots,\frac{3n}{2n};\frac{x^2}{n^n}\right)$$
The name of the function should help you to find more (for example the previous results) :
Haubold, Mathai and Saxena's "Mittag-Leffler Functions and Their Applications".
$$f_a=\sum_{k=0}^\infty\frac{(-1)^kz^{ak}}{\Gamma(1+ak)}$$
If $a$ is a positive integer, asking Wolfram Alpha, you will get the expressions for $a=1,2,3,4$.
For larger values, only hypergeometric functions $$f_a=\, _0F_{a-1}\left(;\frac{1}{a},\frac{2}{a},\cdots,\frac{a-1} {a};-\left(\frac{z}{a}\right)^a\right)$$
