$\theta(z) = \sum_{n=-\infty}^{+\infty} e^{\frac{-n^2}{2}}e^{inz}$ How to show $\theta '(\frac{i}{2}) = \frac{-i}{2} \theta (\frac{i}{2})$?

Question

$$\theta(z) = \sum_{n=-\infty}^{+\infty} e^{\frac{-n^2}{2}}e^{inz}$$

How to show $\theta '(\frac{i}{2}) = \frac{i}{2} \theta (\frac{i}{2})$?

Answer 1

$$\theta'(z)=\sum_n\exp(-n^2/2)ine^{inz}.$$ $$\theta'(i/2)=\sum_n\exp(-n^2/2)ine^{-n/2} =i\sum_n n\exp(-n(n+1)/2).$$ If we pair off the $n$ and the $-1-n$ terms here we get $$\theta'(i/2)=-i\sum_{n=0}^\infty \exp(-n(n+1)/2).$$ But $$(i/2)\theta(i/2)=\frac i2\sum_n\exp(-n(n+1)/2) =i\sum_{n=0}^\infty\exp(-n(n+1)/2).$$