How to find the sum of $\sum_{n=0}^{\infty} \frac{z^{3 n}}{(3 n) !}$?

Question

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?

Answer 1

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.

Answer 2

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