So I am self-learning about Jacobi theta function from Whittaker and Watson's book. In section $\textbf{21.41}$, they introduce the proof for the identity below: $$\vartheta'_1 (0) = \vartheta_2 (0) \vartheta_3 (0) \vartheta_4 (0)$$ I haven't read through it and tried to prove it by myself instead, and here is the proof I came up with:
$\textbf{Part 1:}$ We will show that: $$\vartheta_2 (0) \vartheta_3 (0) \vartheta_4 (0) = 2 q^{1/4} G^3$$ where $G = \prod_{n=1}^{\infty} (1-q^{2n})$ with $ q = e^{i\pi \tau}$. Indeed, by Jacobi triple products, one has: $$\vartheta_2 (0) \vartheta_3 (0) \vartheta_4 (0)=q^{1/4} \left(\sum_{\mathrm{Z}} q^{n^2+n}\right)\left(\sum_{\mathrm{Z}} q^{n^2}\right) \left(\sum_{\mathrm{Z}} (-1)^n q^{n^2}\right) $$$$= q^{1/4}\cdot G^3 \cdot \prod_{n\geq 1} (1+q^{2n-1}\cdot q)(1+q^{2n-1}\cdot q^{-1})(1+q^{2n-1})^2 (1-q^{2n-1})^2$$ $$ = 2q^{1/4}\cdot G^3 \cdot \prod_{n\geq 1}(1+q^{2n})^2\cdot \prod_{n\geq 1}(1-q^{4n-2})^2= 2q^{1/4}G^3$$ The last equality holds because $ \displaystyle \prod_{n\geq 1} (1+z^n) = \prod_{n\geq 1} \frac{1}{1-z^{2n-1}}$ for $z \in \mathbb{C},\vert x \vert < 1$.
$\textbf{Part 2}:$ Knowning that $\vartheta_4 \left(\frac{\pi \tau}{2}\right) = 0$, it is intuitive to compute the derivative of $\vartheta_1$ at $z=0$: $$\vartheta'_1(0) = q^{1/4} \sum_{\mathbb{Z}} (2n+1)q^{n^2+n} (-1)^n = q^{1/4} \left(2\sum_{\mathbb{Z}} n q^{n^2+n}(-1)^n + \sum_{\mathbb{Z}} q^{n^2+n} (-1)^n \right) $$$$= 2q^{1/4} \sum_{\mathbb{Z}} n q^{n^2+n}(-1)^n + q^{1/4} \vartheta_4 \left(\frac{\pi \tau}{2}\right) = 2q^{1/4} \sum_{\mathbb{Z}} n q^{n^2+n}(-1)^n $$ Here, we will try to prove that $\sum_{\mathbb{Z}} n q^{n^2+n}(-1)^n = \frac{1}{2i} \vartheta'_4 \left(\frac{\pi \tau}{2}\right)=G^3$. Again, by Jacobi triple product: $$ \vartheta_4 (z) = G \prod_{n\geq 1} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})$$ $$\Longrightarrow \vartheta'_4 (z)=\vartheta_4 (z) \left(\sum_{n \geq 1} \frac{-2i q^{2n-1}e^{2iz}}{1-q^{2n-1}e^{2iz}} + \frac{2i q^{2n-1}e^{-2iz}}{1-q^{2n-1}e^{-2iz}}\right)$$ $$ \Longrightarrow \lim_{z \to \frac{\pi \tau}{2}} \vartheta'_4 (z)= G^3 \lim_{z \to \frac{\pi \tau}{2}} (1-q e^{-2iz})\cdot \frac{2i}{1- q e^{-2iz}} = 2i G^3$$ Hence: $$ \vartheta'_1 (0) = 2 q^{1/4}\cdot \frac{1}{2i}\vartheta'_4 \left(\frac{\pi \tau}{2}\right) = 2 q^{1/4} G^3 = \vartheta_2 \vartheta_3 \vartheta_4. \square$$ So my question is whether my proof is correct or not? Are there any details that I miss such as convergence, conditions for summation? Thank you so much.