I want to prove $\sum_{n=-\infty}^{\infty} q^{n^2} i^n = \sum_{n=-\infty}^{\infty} q^{4n^2} (-1)^n $

Question

I want to prove \begin{align} \sum_{n=-\infty}^{\infty} q^{n^2} i^n = \sum_{n=-\infty}^{\infty} q^{4n^2} (-1)^n \end{align}

It is quite weird.. even I substitute $n=2k$, then $dn=2$ so we are considering different sums...

This comes from

Answer 1

If $n$ is odd, then $q^{n^2}i^n+q^{(-n)^2}i^{-n}=0$ and therefore\begin{align}\sum_{n=-\infty}^\infty q^{n^2}i^n&=\sum_{\begin{array}{c}n=-\infty\\[-1ex]n\text{ even}\end{array}}^\infty q^{n^2}i^n\\&=\sum_{n=-\infty}^\infty q^{(2n)^2}i^{2n}\\&=\sum_{n=-\infty}^\infty q^{4n^2}(-1)^n.\end{align}

Answer 2

Hint: consider the sum over even $n$ and the sum over odd $n$. Use the fact that $i^{2n+1}=- i^{-(2n+1)}$ so the sum over odd $n$ is $0$.

Answer 3

For $n=2k+1$ we have $$i^n+i^{-n}=i^{2k+1}+i^{-2k-1}=(-1)^k\cdot i+(-1)^k\cdot (-i)=0$$ so $$q^{n^2}i^n+q^{(-n)^2}i^{-n}=0\ \ \text{for all odd } n.$$