Prove $\sum_{n\ge1}\frac{1}{q^n+q^{-n}}=\tfrac14(\vartheta_3^2(q)-1)$

Question

Prove that $$\sum_{n\ge1}\frac{1}{q^n+q^{-n}}=\tfrac14(\vartheta_3^2(q)-1),$$ provided by Wolfam.

Note that here, we use the notational conventions $$\vartheta_3(z,q)=\sum_{n\in\Bbb Z}q^{n^2}e^{2niz},$$ $$\vartheta_3(q)\equiv \vartheta_3(0,q),$$ and of course $\vartheta_3^2(q)=\vartheta_3(q)\cdot\vartheta_3(q)$.

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I have gotten a significant portion of the way. We have that $$f(q)=\sum_{n\ge1}\frac{1}{q^n+q^{-n}}=L(q,-1;q^2),$$ where $$L(a,b;q)=\sum_{n\ge1}\frac{a^n}{1-bq^n}\qquad |q|>1.$$ It can be shown, for sufficient $a$ and $b$, that $$L(a,b;q)=L(b^{-1},a^{-1};q).$$ This is the case here, so we have $$f(q)=L(-1,q^{-1};q^2),$$ which is $$f(q)=L(1,q^{-2};q^4)-L(1,q^{-3};q^4),$$ by splitting the sum up into parts of even and odd index. This may be evaluated in terms of the $q$-digamma function $\psi_q(s)$ as $$f(q)=\tfrac1{4\ln q}\left(\psi_{q^{-4}}(\tfrac34)-\psi_{q^{-4}}(\tfrac14)\right).$$ This is $$4f(q)\ln q=\frac{\partial}{\partial s}\ln\left[\Gamma_{q^{-4}}(\tfrac12+s)\Gamma_{q^{-4}}(\tfrac12-s)\right]\bigg|_{s=1/4}.$$ Then from here we can show that $$\left(\Gamma_{q^{-4}}(\tfrac12+s)\Gamma_{q^{-4}}(\tfrac12-s)\right)^{-1}=\frac{q^{4s^2+3}}{(q^4;q^4)_\infty^3 (1-q^4)}\vartheta_4(-2is\ln q, q^2),$$ but I have no idea about how the logarithmic derivative (w.r.t $s$) of this has anything to do with $\vartheta_3^2$. Could I have some help? Thanks.

Answer 1

There are two ways to prove it, the sum of squares theorem.

• Show that $\Bbb{Z}[i]$ is a PID where $(p)$ is a prime ideal iff $p\equiv 3\bmod 4$, thus $\frac14 \sum_{a,b}(a^2+b^2)^{-s}=\zeta(s)L(s,\chi_4)$

• Show that both $(\sum_n e^{2i\pi n^2 z})^2$ and $1+4\sum_n (\sum_{d| n}\chi_4(d)) e^{2i\pi nz}\in M_1(\Gamma_1(4))$ and that the latter space of modular forms is finite dimensional thus it suffices to compare finitely many coefficients to prove they are equal