Identities for power series like $\sum_n z^{n^3}$

Question

Probably, one of the first power series that every mathematician encounter is the geometric series

$$\sum_{n=0}^\infty z^n = \frac1{1-z}, \quad z \in \mathbb{C},\; |z| < 1 .$$

Also, a particular case of the Jacobi triple product gives a beautiful identity for the power series with $z$ raised to the square numbers

$$\sum_{n=-\infty}^\infty z^{n^2} = \prod_{m=1}^\infty \left(1-z^{2m}\right)\left(1+z^{2m-1}\right)^2, \quad z \in \mathbb{C},\; |z| < 1 .$$ However, I have never seen an equally beautiful identity involving $\sum_n z^{n^3}$. Clearly, such series appear, for example, in papers regarding the positive integers which are sum of $k$ cubes, since the fact that $(\sum_{n=0}^\infty z^{n^3})^k = \sum_{n=0}^\infty r_k(n) z^n$, where $r_k(n)$ is the number of ways to write $n$ as a sum of $k$ cubes, can be exploited. But this is a general generating-functions property that has nothing special to do with the cubes.

So my question is: Are there known other nice identities involving series similar to $\sum_n z^{n^k}$, for some integer $k \geq 3$?

Answer 1

One nice relation can be written as follow: $$\sum\limits_{n = - \infty }^ \infty z^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n z^{n^3}= \sum\limits_{n = - \infty }^ \infty (-1)^n z^{2n^3} \phi(z^{6n}) $$

$\quad z \in \mathbb{C},\; |z|=1 $

Where $$\phi(q)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{m^2}$$

More generally if we write $$\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}= \sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m} z^{2n} q^{2(n^2+m^2)} h^{2n(n^2+3m^2)} \tag{1} $$

The proof of relation (1):

$$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

$$F(-z)=\sum\limits_{n = - \infty }^ \infty (-z)^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-z)^n q^{n^2} h^{n^3}$$

$$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-1)^n z^n q^{n^2} h^{n^3}$$ $$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

Because of $F(-z)=F(z)$ , we can write that:

$$F(z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)$$

We need to find $A_n(q,h)$

$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

Let's use the transformation

$z=ZQ^{2}h^{3}$

$q=Qh^{3}$

$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n} Q^{2n^2+4n} h^{2n^3+6n^2+6n} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2+2n} h^{n^3+3n^2+3n} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^n Q^{n^2+2n} h^{n^3+3n^2+3n}$$

If we multiply both side by $Z^2Q^2h^2$

$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n+2} Q^{2n^2+4n+2} h^{2n^3+6n^2+6n+2} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1}$$

$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2(n+1)} Q^{2(n+1)^2} h^{2(n+1)^3} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{(n+1)^2} h^{(n+1)^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{(n+1)^2} h^{(n+1)^3}$$

$$\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{2n} Q^{2n^2} h^{2n^3} A_{n-1}(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n} Q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{n} Q^{n^2} h^{n^3}$$

$$\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{2n} q^{2n^2} h^{2n^3} A_{n-1}(qh^3,h)=\sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{n} q^{n^2} h^{n^3}$$

$$A_n(q,h)=A_{n-1}(qh^3,h)$$ $$A_{-1}(q,h)=A_{0}(qh^{-3},h)$$ $$A_1(q,h)=A_{0}(qh^3,h)$$

$$A_2(q,h)=A_{1}(qh^3,h)=A_{0}(qh^6,h)$$ $$A_{-2}(q,h)=A_{-1}(qh^{-3},h)=A_{0}(qh^{-6},h)$$ $$A_3(q,h)=A_{2}(qh^3,h)=A_{1}(qh^6,h)=A_{0}(qh^9,h)$$ $$A_n(q,h)=A_{0}(qh^{3n},h)$$

We need to find $A_{0}(q,h)$ to complete the proof:

We need to focus on $z^0$ terms

$$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

If we multiply terms by terms for $z^0$ terms

$$A_{0}(q,h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}$$ $$A_{n}(q,h)=A_{0}(qh^{3n},h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}$$

Thus We can write that

$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

$$\sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m}z^{2n} q^{2n^2+2m^2} h^{2n^3+6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \phi(q^2h^{6n}) =\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$

If we put that $z=1$ and $q=1$

$$\sum\limits_{n = - \infty }^ \infty (-1)^n h^{2n^3} \phi(h^{6n}) =\sum\limits_{n = - \infty }^ \infty h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n h^{n^3} $$

Finally, the same method can be used for $\sum\limits_{n = - \infty }^ \infty z^{n^k}$ to get similiar identies where $k>3$