We know the power series representation of #1/(1-x) = sum_nx^n AAx# such that #absx < 1#. So #1/(1+x^2) = (arctan(x))' = sum_n (-1)^nx^(2n)#.
So the power series of #arctan(x)# is #intsum_n (-1)^nx^(2n)dx = sum_n int(-1)^nx^(2n)dx = sum_n((-1)^n)/(2n+1)x^(2n+1)#.
You divide it by #x#, you find out that the power series of #arctan(x)/x# is #sum_n((-1)^n)/(2n+1)x^(2n)#. Let's say #u_n = ((-1)^n)/(2n+1)x^(2n)#
In order to find the radius of convergence of this power series, we evaluate #lim_(n -> +oo)abs((u_(n+1))/u_n#.
#(u_(n+1))/u_n = (-1)^(n+1)*x^(2n+2)/(2n+3)(2n+1)/((-1)^nx^(2n)) = -(2n+1)/(2n+3)x^2#.
#lim_(n -> +oo)abs((u_(n+1))/u_n) = abs(x^2)#. So if we want the power series to converge, we need #abs(x^2) = absx^2 < 1#, so the series will converge if #absx <1#, which is not surprising since it's the radius of convergence of the power series representation of #arctan(x)#.
