How do I evaluate this sum :$\sum_{n=1}^{\infty}\frac{{(-1)}^{n²}}{{(i\pi)}^{n}}$?

Question

I'm interesting to know how do i evaluate this sum :$$\sum_{n=1}^{\infty}\frac{{(-1)}^{n²}}{{(i\pi)}^{n}}$$, I have tried to evaluate it using two partial sum for odd integer $n$ and even integer $n$ ,but i can't since it's alternating series ,and i would like to know if it's well know series also what about it's values :real or complex ? .

Note : wolfram alpha showed that is a convergent series by the root test

Thank you for any help

Answer 1

$(-1)^{n^2}=(-1)^n$ and $\displaystyle \sum_{n=1}^{\infty}\left(-\frac{1}{i\pi}\right)^n$ is a geometric series. Since $\left|\frac{1}{i\pi}\right|=\frac{1}{\pi} < 1$, The series converges to $\frac{-\frac{1}{\pi i}}{1+\frac{1}{\pi i}}$.

Answer 2

$$\sum_{n=1}^{\infty}\frac{{(-1)}^{n²}}{{(i\pi)}^{n}} =\sum_{n=1}^{\infty}\frac{{(-1)}^{n}}{{(i\pi)}^{n}} = \sum_{n=1}^{\infty}\frac{{1}}{{(-i\pi)}^{n}}$$

voilà geometric series!