Finding singularities in circle of convergence of $f(z)$ and showing taylor series diverges there $$f(x)=\arctan(x)$$
I know that $$\dfrac{1}{1+x^2}=1-x+x^2-x^3+\cdots=\sum_{i=0}^\infty (-x)^i$$
Also:  $$\arctan(x)=\int \dfrac{1}{1+x^2}dx = \int \sum_{i=0}^\infty (-x)^i dx = x - 0.5x^2+1/3x^3+\cdots=\sum_{i=0}^\infty \frac{(-1)^i(x)^{i+1}}{2i+1}$$
Radius of convergence $R = 1$ using ratio test $\implies |z|\le1$ is the circle of convergence
Singularities are at $\pm i$ and both lie in $|z|$ however how do i show that the series diverges there?
EDIT:
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +\cdots + (-1)^kx^{2k} + \cdots\to \int \dfrac{dx}{1+x^2} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}.$$ 

If a power series converges in some point $x_0$, then the derivation also converges
 in the point $x_0$. Here, the derivation diverges at $x_0=-1$ and $x_0=1$, so the
 original power series diverges there as well.
