Start from the sum of the geometric series:
sum_(n=0)^oo q^n = 1/(1-q)∞∑n=0qn=11−q
for abs q <1|q|<1.
Let q = -x^2q=−x2:
1/(1+x^2) = sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n)11+x2=∞∑n=0(−x2)n=∞∑n=0(−1)nx2n
converging for abs x <1|x|<1.
In the interval of convergence we can integrate term by term:
int_0^t dx/(1+x^2) = sum_(n=0)^oo (-1)^n int_0^t x^(2n)dx∫t0dx1+x2=∞∑n=0(−1)n∫t0x2ndx
arctan t = sum_(n=0)^oo (-1)^n t^(2n+1)/(2n+1)arctant=∞∑n=0(−1)nt2n+12n+1
Let now t=x^3t=x3:
arctan x^3 = sum_(n=0)^oo (-1)^n (x^3)^(2n+1)/(2n+1)arctanx3=∞∑n=0(−1)n(x3)2n+12n+1
arctan x^3 = sum_(n=0)^oo (-1)^n x^(6n+3)/(2n+1)arctanx3=∞∑n=0(−1)nx6n+32n+1
and truncating at the third term:
arctan x^3 = x^3-x^9/3+x^15/5+o(x^15)arctanx3=x3−x93+x155+o(x15)