Question

What is the result?
(-1)^(n^2/2) = 1if n is even
what is the result if n is odd?
Please, help

Answer 1

@dan815

Answer 2

try it

Answer 3

Can you?

Answer 4

This is the original problem:
Find the radius convergence of \(\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}\)

Answer 5

We need an argument for i^n^2

Answer 6

To be fair the root of \(-1\) can be \(+i\) or \(-i\).

Answer 7

n=even
that thing is 1

n=odd
then its i

Answer 8

wait n=odd has got cases

Answer 9

Exactly, we can't tell if it's \(i\) or \(-i\) right?

Answer 10

\(i^{n^2}= (i^2)^{n^2/2}=(-1)^{n^2/2}\)

Answer 11

yea tru

Answer 12

ok, then , right?