Infinite sum and Jacobi Theta function

Question

I have encountered a sum as following:

\begin{equation} \sum_{n=1}^{\infty} q^{n^2} \end{equation}

\begin{equation} \sum_{n=1}^{\infty} n^2 q^{n^2} \end{equation}

where $0<q<1$.
I know that the first sum is related to Jacobi Theta function, but what about second sum? Can I do anything about that?

Answer 1

In the realm of the functions used herein, different conventions exist in what to use for $q$ and what differentiation operator to use. Therefore, let me do some declarations first.

Notation: Lattices in $\mathbb{C}$ are characterized by their complex period ratio $\tau$ which is taken such that its imaginary part is positive. Define $q = \exp(2\pi\mathrm{i}\tau)$ and more generally $q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$ for positive integer $n$. We will interpret expressions using $q$ or some $q_n$ as being functions of $\tau$.

I will use the differentiation operator $$\dot{(\ )} = \frac{1}{2\pi\mathrm{i}}\frac{\mathrm{d}(\ )}{\mathrm{d}\tau} = q\frac{\mathrm{d}(\ )}{\mathrm{d}q} = \frac{1}{n}q_n\frac{\mathrm{d}(\ )}{\mathrm{d}q_n}$$

The $q$ used in the question would be written $q_2$ in this answer.

Some related well known functions:

• Dedekind eta function: $$\eta(\tau) = q_{24}\prod_{n=1}^\infty(1 - q^n) = \sum_{k\in\mathbb{Z}} (-1)^k q_{24}^{(6k-1)^2}$$
• Jacobi thetanulls: $$\begin{align} \Theta_{00}(\tau) &= \frac{\eta^2\!\left(\frac{\tau+1}{2}\right)}{\eta(\tau+1)} = \sum_{k\in\mathbb{Z}} q_2^{k^2}, & \text{aka $\theta_3(0,q_2)$,} \\\Theta_{01}(\tau) &= \frac{\eta^2\!\left(\frac{\tau}{2}\right)}{\eta(\tau)} = \sum_{k\in\mathbb{Z}} (-1)^k q_2^{k^2}, & \text{aka $\theta_4(0,q_2)$,} \\\Theta_{10}(\tau) &= \frac{2\eta^2(2\tau)}{\eta(\tau)} = \sum_{k\in\mathbb{Z}} q_8^{(2k+1)^2}, & \text{aka $\theta_2(0,q_2)$,} \end{align}$$
• Eisenstein series $\operatorname{E}_2(\tau)$, also known as Ramanujan's $P(q)$: $$\begin{align} \operatorname{E}_2(\tau) &= P(q) = 24\frac{\dot{\eta}(\tau)}{\eta(\tau)} = 1 - 24\sum_{n=1}^\infty \frac{n\,q^n}{1 - q^n} = 1 - 24\sum_{n=1}^\infty \sigma(n)\,q^n \end{align}$$ where $\sigma(n) = \sum_{d\mid n}d$ is the divisor sum.

Answer: You look for an identity involving $$\frac{1}{2}q_2\frac{\mathrm{d}}{\mathrm{d}q_2}\left(\theta_3(0,q_2) - 1\right) = \dot{\Theta}_{00}(\tau)$$

• Since the Jacobi thetanulls are eta quotients, their logarithmic derivatives can be expressed using $\operatorname{E}_2$: $$\begin{align} \frac{\dot{\Theta}_{00}(\tau)}{\Theta_{00}(\tau)} &= \frac{1}{24}\left(\operatorname{E}_2\!\left(\frac{\tau+1}{2}\right) - \operatorname{E}_2(\tau)\right) = \frac{1}{24}\left(P(-q_2) - P(q)\right) \\&= \sum_{n=1}^\infty (-1)^{n+1} \sigma_{\bar{\mathrm{u}}}(n)\,q_2^n \qquad\text{(power series)} \\&= \sum_{n=1}^\infty \frac{(-1)^{n+1}n\,q_2^{n}}{1 - q_2^{2n}} \qquad\text{(Lambert-like series)} \\&= \sum_{n=1}^\infty \frac{q_2^{2n-1}}{\left(1 + q_2^{2n-1}\right)^2} \qquad\text{(transformed)} \\\text{where}\quad \sigma_{\bar{\mathrm{u}}}(n) &= \sum_{\substack{d\mid n\\2d\nmid n}} d \qquad\text{(sum of divisors whose cofactors are odd)} \end{align}$$ Above, transformed means: Express the summand fractions as power series, swap the summation's nesting order, and simplify the new inner series.
• Such logarithmic derivatives of Jacobi thetanulls are known as Halphen's Psi functions. Define $$\begin{align} \Psi_{00}(\tau) &= \frac{\dot{\Theta}_{00}(\tau)}{\Theta_{00}(\tau)} &\Psi_{01}(\tau) &= \frac{\dot{\Theta}_{01}(\tau)}{\Theta_{01}(\tau)} &\Psi_{10}(\tau) &= \frac{\dot{\Theta}_{10}(\tau)}{\Theta_{10}(\tau)} \end{align}$$ There are a lot of identities relating the Jacobi thetanulls, and those can be translated into identities for the Psi functions. Furthermore, series representations for the Psi functions can be obtained via $\operatorname{E}_2$, as shown above for $\Psi_{00}$. Related OEIS sequences: A186690, A002131, A002129. I will not cover the details here. However, some aspects are particularly interesting, so I will mention them here.
• Pairwise differences of Psi functions can be expressed in terms of thetanulls: $$\begin{align} \Psi_{10} - \Psi_{01} &= \frac{\Theta_{00}^4}{8} &\Psi_{10} - \Psi_{00} &= \frac{\Theta_{01}^4}{8} &\Psi_{00} - \Psi_{01} &= \frac{\Theta_{10}^4}{8} \end{align}$$ Consequently, when differentiating a quotient of two Jacobi thetanulls, the result can be expressed without Psi functions. Example: $$\left(\frac{\Theta_{10}}{\Theta_{00}}\right)^{\cdot} = (\Psi_{10} - \Psi_{00})\frac{\Theta_{10}}{\Theta_{00}} = \frac{\Theta_{01}^4}{8}\frac{\Theta_{10}}{\Theta_{00}}$$ This also implies that we can express $\Theta_{00}^4, \Theta_{01}^4,\Theta_{10}^4$ in terms of the Eisenstein series $\operatorname{E}_2$.
• For further differentiation, we need to know how to express derivatives of the Psi functions. To cut things short, the result is $$\begin{align} \dot{\Psi}_{00} &= 2\left( \Psi_{00}\,\Psi_{01} + \Psi_{00}\,\Psi_{10} - \Psi_{01}\,\Psi_{10}\right) \\\dot{\Psi}_{01} &= 2\left( \Psi_{00}\,\Psi_{01} + \Psi_{01}\,\Psi_{10} - \Psi_{00}\,\Psi_{10}\right) \\\dot{\Psi}_{10} &= 2\left( \Psi_{00}\,\Psi_{10} + \Psi_{01}\,\Psi_{10} - \Psi_{00}\,\Psi_{01}\right) \end{align}$$ An equivalent formulation is $$\begin{align} \dot{\Psi}_{01} + \dot{\Psi}_{10} &= 4\,\Psi_{01}\,\Psi_{10} \\\dot{\Psi}_{00} + \dot{\Psi}_{10} &= 4\,\Psi_{00}\,\Psi_{10} \\\dot{\Psi}_{00} + \dot{\Psi}_{01} &= 4\,\Psi_{00}\,\Psi_{01} \end{align}$$
• Whenever you run into expressions crowded with $\operatorname{E}_2$ and its derivatives, Halphen's Psi functions offer nicely symmetric expressions for substitution: $$\begin{align} \frac{\operatorname{E}_2}{8} &= \Psi_{00} + \Psi_{01} + \Psi_{10} \\\frac{\dot{\operatorname{E}}_2}{16} &= \Psi_{00}\,\Psi_{01} + \Psi_{00}\,\Psi_{10} + \Psi_{01}\,\Psi_{10} \\\frac{\ddot{\operatorname{E}}_2}{192} &= \Psi_{00}\,\Psi_{01}\,\Psi_{10} \end{align}$$
• Applying the above, you will find that some higher derivatives can be expressed without Psi functions. Example: $$\left(\frac{1}{\Theta_{00}^2}\right)^{\cdot\cdot} = -\frac{1}{16}\,\frac{\Theta_{01}^4\,\Theta_{10}^4}{\Theta_{00}^2}$$ You are invited to find the second derivatives of $\Theta_{01}^{-2}$ and $\Theta_{10}^{-2}$ as an exercise.

I hope this helps.

Answer 2

A series expansion of an Elliptic Theta function (up to order 100) is:

$1-2 q+2 q^4-2 q^9+2 q^{16}-2 q^{25}+2 q^{36}-2 q^{49}+2 q^{64}-2 q^{81}+2 q^{100}+O\left(q^{101}\right)$

Answer 3

For the first sum, we have a Jacobi Theta function, just as you claim.

$$\vartheta_3(q)=\sum_{n=-\infty}^\infty q^{n^2}\\\frac12(\vartheta_3(q)-1)=\sum_{n=1}^\infty q^{n^2}$$

Take the derivative w.r.t. $q$ and you will find that

$$\frac12\vartheta_3'(q)=\sum_{n=1}^\infty n^2q^{n^2-1}$$

$$\frac q2\vartheta_3'(q)=\sum_{n=1}^\infty n^2q^{n^2}$$

Though one cannot improve on this form, according to Wolfram. Note these series representations themselves converge extremely fast for $|q|<1$, so I'm not quite sure on any forms that would converge faster.

But anyways, if that's what you want,

$$\sum_{n=1}^\infty n^2q^{n^2}=\sum_{n=1}^\infty\left[4n^2(q^4)^{n^2}+(-1)^{n+1}n^2q^{n^2}\right]$$

$$\frac q2\vartheta_3'(q)=2q^4\vartheta_3'(q^4)-\frac q2\vartheta_4'(q)$$

Repeatedly apply this and you will get

$$\frac q2\vartheta_3'(q)=-\sum_{n=0}^\infty2^{2n-1}q^{4^{n-1}}\vartheta_4'(q^{4^n})$$