Taylor expansion of arctan x How can I prove that the tailor expansion of arctan x converges only at [-1,1]? I was able to prove that it converges at the given interval but I don't know how to prove that it does not converge anywhere else . Can someone help?

The derivative of $\arctan x$ is
$$\frac d{dx} \arctan x =\frac 1{1+x^2}$$
From the formula for geometric series, we have that
$1+y+y^2+y^3+...=\frac 1{1−y}$ if and only if $|y|<1$.
Plugging in $-x^2$ for $y$, we get that 
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; i.e., provided that $|x|\lt 1$.
And if $|x|\lt 1$, then $x \in [-1,1]$. If $x$ isn't in this interval, then we do not get convergence.
