Evaluate $\sum_{n=0}^{\infty} \frac{z^{kn+p}}{(kn+p)!}$ $$\sum_{n=0}^{\infty} \frac{z^{kn+p}}{(kn+p)!}$$
I have found out that 
$$\sum_{n=0}^{\infty} \frac{z^{kn}}{(kn)!} = \frac{1}{k} {}\sum_{i=0}^{k-1} e^{z\varepsilon_i}$$
where $\varepsilon_i$ is the $i$-th rooth of unity.
how can this help further, I tried calculating by multiplying with every term  $e^{z\varepsilon_i}$ with $ \varepsilon_i^s z^s $ and then adding them togheter but I'm having troubles on how to get the $ (kn+p) $! on the denominator.
 A: Hint:
Integrate both sides of your equation $ p$ times from $0$ to $z'$ over $z$ (after each integration replace $z'$ with $z$).

The result is: 
$$\sum_{n=0}^{\infty} \frac{z^{kn+p}}{(kn+p)!} = 
\frac{1}{k}\sum_{i=0}^{k-1}\frac1{\varepsilon_i^p}\left(e^{\varepsilon_iz}-\sum_{q=0}^{p-1}\frac{(\varepsilon_i z)^q}{q!}\right)\tag1$$
A closer look at the expression in parentheses reveal that it is the exponent of $\varepsilon_i z$ without first $p$ terms of its Maclaurin series.
The rhs of (1) can be further simplified to
  $$\frac{1}{k}\sum_{i=0}^{k-1}\frac{e^{\varepsilon_iz}}{\varepsilon_i^p}-\sum_{i=1}^{\left\lfloor\frac pk\right\rfloor}\frac{z^{p-ki}}{(p-ki)!}$$
  since $\frac1k\sum_{i=0}^{k-1}{\varepsilon_i^{q-p}}=\delta_{q\operatorname{mod}k,p}.$

The expression  (1) can be easily proved by induction. Let 
$$f_p (z)=\sum_{n=0}^{\infty} \frac{z^{kn+p}}{(kn+p)!},\quad 
F_p (z)=\frac{1}{k}\sum_{i=0}^{k-1}\frac1{\varepsilon_i^p}\left(e^{\varepsilon_iz}-\sum_{q=0}^{p-1}\frac{(\varepsilon_i z)^q}{q!}\right)$$
As you have proved:
$$f_0(z)=F_0 (z).$$
Further:
$$ f_p(z)=F_p (z)\implies f_{p+1}(z)=F_{p+1}(z).$$
Indeed:
$$ f_{p+1}(z)=\int_0^zf_p(z')dz'\stackrel {I.H.}=
\int_0^z F_p (z')dz'=F_{p+1}(z).$$
A: In general, let $f(z)$ be a power series, say
$$f(z)=\sum_{n=0}^\infty a_nz^n,$$
where $a_0,a_1,a_2,\ldots$ are complex numbers.
Then define
$$f_{k,p}(z)=\sum_{n=0}^\infty a_{kn+p}z^{kn+p}$$
for every positive integer $k$ and for any non-negative integer $p$. Note that when $p=0,1,2,\ldots,k-1$, $f_{k,p}(z)$ is given by
$$f_{k,p}(z)=\frac{1}{k}\sum_{r=0}^{k-1}e^{-\frac{2rp\pi i}{k}}f\left(e^{\frac{2r\pi i}{k}}z\right).$$
This is due to the following identity: for any integer $t$,
$$\frac{1}{k}\sum_{r=0}^{k-1} e^{\frac{2rt\pi i}{k}}=\left\{\begin{array}{ll}1&\text{if }k\mid t,\\0&\text{otherwise}.\end{array}\right.$$
If $p\geq k$, I don't think there is a nice way to simplify $f_{k,p}$ except through truncation:
$$f_{k,p}(z)=f_{k,s}(z)-\sum_{n=0}^{q-1}a_{kn+s}z^{kn+s}$$
if
$$p=kq+s$$
for some integers $q$ and $s$ such that $q\geq 1$ and $0\leq s<k$.
