Problem: Find a "Closed form" for the given power series:
\begin{align*} \sum_{n=1}^{\infty} z^{3n}.\end{align*}
Answer:
Since we know that
\begin{align*} \sum_{n=0}^{\infty} z^{n} &= \frac{1}{1-z}\\
\sum_{n=0}^{\infty} (z^3)^n &= \frac{1}{1-z^3}\end{align*}
Thus,
\begin{align*}
\sum_{n=1}^{\infty} z^{3n} &= \sum_{n=0}^{\infty} z^{3n} - z^{3*0}\\
&=\sum_{n=0}^{\infty} z^{3n} - 1\\
&= \frac{1}{1-z^3} - 1 \\
\end{align*}
Therefore, the closed form for $ \sum_{n=1}^{\infty} z^{3n}$ is $\frac{1}{1-z^3} - 1 $.