calc

u guys know how to get the sum of geometric series from 0 to infinity of (-1)^n * x^2n

bro don't u have a piazza u could ask?

the sum of an infinite series is first term/ (1-r)

r= the thing you multiply by, basically find the first and second term, divide second by first and you have r then just put it into the formula

1/1+x^2

Looking at the first term, n=1, we have:
(-1)^n x^2n = (-1)^1 x^2(1) = -(x^2)
Then 2nd term is:
(-1)^2 x^2(2) = x^4
Then 3rd term is -(x^6)
So we can see that n=1, or the a-value, is -(x^2), and the common ratio is -(x^2), since it goes up by a power of 2 each time, and flips the sign.

Like akira said, the sum of an infinite geometric series is a/(1-r), so here we have:
Sum = -(x^2)/(1- -(x^2))
= - (x^2/1+x^2)