1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

viola geometric series!

$\endgroup$
1
$\begingroup$

$(-1)^{n^2}=(-1)^n$ 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 $\frac{-\frac{1}{\pi i}}{1+\frac{1}{\pi i}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.