Finding taylor series

use summation from 0 to infinity of (-1)^n (x^2)^n to find a taylor series for arctan x around x =0

Note that, since:
sum(n=0 to infinity) [(-1)^n * (x^2)^n]
= sum(n=0 to infinity) (-x^2)^n
= 1/[1 - (-x^2)], by the infinite geometric series
= 1/(1 + x^2),

and the integral of 1/(1 + x^2) is arctan(x), we can just integrate this series to yield:
arctan(x) = sum(n=0 to infinity) [(-1)^n * x^(2n + 1)]/(2n + 1).
(Note that (x^2)^n = x^(2n) and, so, the integral of x^(2n) is x^(2n + 1)/(2n + 1).)

I hope this helps!