Hello !
- The Maclaurin serie for #\text{arctan}(t)# is
#\text{arctan}(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{2n+1}#.
Proof. #\text{arctan}'(t) = \frac{1}{1+t^2} = (1 - (-t^2))^{-1}# and we know that #(1-X)^{-1} = \sum_{n=0}^\infty X^n# (for #|X|<1#). Actually,
#\text{arctan}'(t) = \sum_{n=0}^\infty (-t^2)^n = \sum_{n=0}^\infty (-1)^n t^{2n}#
After integration, you get
#\text{arctan}(t)-\text{arctan}(0) = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{2n+1}#
The proof is ended because #\text{arctan}(0)=0#.
- Because #x^3\to 0# when #x\to 0#, you can compose and write
#\text{arctan}(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1}#
If you don't like #\sum# symbol,
#\text{arctan}(x^3) = x^3 - \frac{x^9}{3} + \frac{x^15}{5} - \frac{x^{21}}{7} + \ldots #
