How to find the sum of $\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}$? I'm really stuck with this problem and I hope some of you could give me a hint.
Consider the functions: $$f(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}, \quad f^{\prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}, \quad f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}, \quad f^{\prime \prime \prime}(z)=f(z)$$
I need to find the sum of each of these series.
My first thought was to show that:
$$f(z)+f^{\prime}(z)+f^{\prime \prime}(z)=\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+2}}{(3 n+2) !}+\sum_{n=0}^{\infty} \frac{z^{3 n+1}}{(3 n+1) !}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !}=e^{z}$$
And then solve this differential equation.
However, officially I haven't had a course in differential equations. So is there another way I can solve this using basic Complex Analysis? I hope some of you can guide me in the right direction, what theorems could be useful?
 A: In general, it is possible for us to extract components of power series by playing with roots of unity. Particularly, let's say
$$
f(z)=\sum_{n=0}^\infty c_nz^n
$$
converges absolutely for some $z\in\mathbb C$. If we set $\zeta=e^{2\pi i/q}$ then
$$
\frac1q\sum_{r=0}^{q-1}\zeta^{-ra}f(\zeta^rz)=\sum_{k=0}^\infty c_{qk+a}z^{qk+a}\tag1
$$
Consequently, we get
$$
\sum_{k=0}^\infty{z^{3k+1}\over(3k+1)!}=\frac13\sum_{r=0}^2 e^{-2\pi ir/3}\exp(e^{2\pi ir/3}z)
$$
Proof for (1):
Given $f$ converges absolutely at $z$, we can plug in the power series expansion for $f(z)$ to get
$$
\begin{aligned}
\sum_{r=0}^{q-1}\zeta^{-ra}\sum_{n=0}^\infty c_n\zeta^{rn}z^n
&=\sum_{n=0}^\infty c_nz^{n}\left(\sum_{r=0}^{q-1}\zeta^{r(n-a)}\right)
\end{aligned}
$$
If $n\equiv0\pmod q$, then
$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}=\sum_{r=0}^{q-1}1=q
$$
Otherwise we have $\zeta^{n-a}\ne1$:
$$
\sum_{r=0}^{q-1}\zeta^{r(n-a)}={\zeta^{q(n-a)}-1\over\zeta^{n-a}-1}=0
$$
Consequently, the purpose of sum of roots of unity in (1) is to extract coefficients from power series.
A: Let $\omega$ be the cube root of unity then denote $f_1$ by $e^{\omega x}$,$f_2$ by $e^{\omega^2 x}$ and $f_3$ by $e^x$
$$\begin{align*}f_1 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \omega \dfrac{x^{3n+1}}{(3n+1)!} + \omega^2 \dfrac{x^{3n}}{(3n+2)!}  \right)\\
f_2 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \omega^2 \dfrac{x^{3n+1}}{(3n+1)!} + \omega \dfrac{x^{3n+2}}{(3n+2)!}  \right)\\
f_3 &= \sum_{n\ge0} \left(\dfrac{x^{3n}}{(3n)!} + \dfrac{x^{3n+1}}{(3n+1)!} +  \dfrac{x^{3n}}{(3n+2)!}  \right) \end{align*} $$
Add all of them and get your result
