Is there a closed form expression or an $\textbf{approximation}$ for the below summation:
$\sum_{x=0}^{\infty} \frac{\lambda^{kx}}{{(kx)!}}$ where k is a constant.
I know that if $k=1$, it reduces to $e^{x}$. But what happens for general values of $k$.
On
To elaborate on G Cab’s answer: your series can be expressed as $$f_k(\lambda)=\sum_{n=0}^\infty \frac{\lambda^{kn}}{(kn)!}=\frac{1}{k}\sum_{n=0}^{k-1} e^{\lambda \zeta_k^n}$$ where $\zeta_k$ is the kth root of unity defined as $\zeta_k:=e^{2\pi i/k}$.
To give a few examples: $$f_2(x)=\frac{e^x+e^{-x}}{2}$$ $$f_3(x)=\frac{e^x+e^{\sqrt 3 x/2}\cos(x/2)+e^{-\sqrt 3x/2}\cos(x/2)}{3}$$ $$f_4(x)=\frac{e^x+e^{-x}+2\cos(x)}{4}$$
That leads to an interesting family of functions, akin to $\cosh$ , $\sinh$ ( I do not know if they have a standard name).
In fact, changing a bit the symbols and putting $$ e^{\,z} = \sum\limits_{0\, \le \,n} {{{z^{\,n} } \over {n!}}} = \sum\limits_{\scriptstyle 0\, \le \,j \atop \scriptstyle 0\, \le \,k\; \le \,h} {{{z^{\,j\,\left( {h + 1} \right) + k} } \over {\left( {j\,\left( {h + 1} \right) + k} \right)!}}} = \sum\limits_{0\, \le \,l\; \le \,h} {{\rm cemh}_{\;h,\,k} (z)} $$ for the first values of $h$ we get $$ \eqalign{ & h = 0\quad \Rightarrow \quad {\rm cemh}_{\;0,\,0} (z) = e^{\,z} \cr & h = 1\quad \Rightarrow \quad \left\{ \matrix{ {\rm cemh}_{\;1,\,0} (z) = \cosh (z) \hfill \cr {\rm cemh}_{\;1,\,1} (z) = \sinh (z) \hfill \cr} \right. \cr} $$
So these families are a decomposition of $e^z$ into modular powers, continuing from the even/odd decomposition given by $\cosh, \sinh$.
It is easy to demonstrate that they share many properties of the hyperbolic trig function, starting from that the derivative / integral cycles on the second index $$ \int {{\rm cemh}_{\;h,\,n} (z)dz} = {\rm cemh}_{\;h,\,\,\bmod (n + 1,h + 1)} (z) $$
Also, through the theory of formal power series, it is possible to express them through a combination of the exp of the unit roots.
--- addendum ---
in Wikipedia it is called multisection of a power series