Answer
Diverges for $|x| \gt 1$
Work Step by Step
The Taylor series for $\tan^{-1} x $ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....; |x| \leq 1$
We need to apply the ratio test.
$\lim\limits_{n \to \infty}|\dfrac{u_{n+1}}{u_n}|=\lim\limits_{n \to \infty} |\dfrac{(-1)x^2 (2n-1)}{2n+1}| =x^2 \lim\limits_{n \to \infty} \dfrac{2-1/n1}{2+1/n}=x^2$
When $ x=0.1$, we have:
$| Error|=|\dfrac{(-1)^n x^n}{n}|=\dfrac{1}{n (10^n)}$
But $\dfrac{1}{2n-1} \lt 10^{-3} \implies \dfrac{1}{2n-1} \lt \dfrac{1}{10^{3}} $
or, $2n \gt 1001 \implies n=501$
So, the series diverges for $|x| \gt 1$.
