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:
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +\cdots + (-1)^kx^{2k} + \cdots\to \int \dfrac{dx}{1+x^2} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}.$$ 

$$
\frac 1 {1+x^2} = \frac 1 {(x-i)(x+i)} = \frac {i/2} {x+i} - \frac{i/2}{x-i}
$$
In a small neighborhood of $i$ this is a continuous function plus $\text{constant}/(x-i)$.
Think of what happens to a function whose derivative is $1/(x-i)$ as $x$ approaches $i$ from the positive real direction.  If $x$ is approaching from that direction, then $x-i$ is real, and the antiderivative of $1/(x-i)$ at least on the ray going in that direction from $i$ will be a constant plus $\log(x-i)$.  That function has a pole.
