How to derive the gregory series for inverse tangent function? How to derive the Gregory series for the inverse tangent function?
Why is Gregory series applicable only to the set $ [-\pi/4,\pi/4] $ ?

You need to know that $$\arctan x = \int_{0}^{x}\frac{dt}{1 + t^{2}}\tag{1}$$ In fact this equation can be used as a starting point to develop the theory of circular functions. But let's not worry about that here.
We know (via the formula for sum of a geometric progression) that $$1 - t^{2} + t^{4} - t^{6} + \cdots + (-1)^{n - 1}t^{2n - 2} = \frac{1 - (-1)^{n}t^{2n}}{1 + t^{2}}\tag{2}$$ and therefore $$\frac{1}{1 + t^{2}} = 1 - t^{2} + t^{4} - t^{6} + \cdots + (-1)^{n - 1}t^{2n - 2} + (-1)^{n}\cdot\frac{t^{2n}}{1 + t^{2}}\tag{3}$$ and hence on integrating the above equation between $0$ and $x$ and using $(1)$ we get $$\arctan x = x - \frac{x^{3}}{3} + \cdots + (-1)^{n - 1}\cdot\frac{x^{2n - 1}}{2n - 1} + (-1)^{n}\int_{0}^{x}\frac{t^{2n}}{1 + t^{2}}\,dt$$ or $$\arctan x = x - \frac{x^{3}}{3} + \cdots + (-1)^{n - 1}\cdot\frac{x^{2n - 1}}{2n - 1} + (-1)^{n}R_{n}(x)\tag{4}$$ where $$R_{n}(x) = \int_{0}^{x}\frac{t^{2n}}{1 + t^{2}}\,dt\tag{5}$$ Now from the above equation it is easy to check that $R_{n}(-x) = -R_{n}(x)$. We will prove that if $0 \leq x \leq 1$ then $R_{n}(x) \to 0$ as $n \to \infty$. Since $R_{n}(-x) = -R_{n}(x)$ it will imply that the $\lim_{n \to \infty}R_{n}(x) = 0$ holds for all $x$ with $|x|\leq 1$.
Now let's assume that $0 \leq x \leq 1$. Then we know that $$\frac{t^{2n}}{1 + t^{2}} \leq t^{2n}$$ for all $t$ with $0 \leq t \leq x$. Hence on integrating the above between $0$ and $x$ we get $$0 \leq R_{n}(x) \leq \int_{0}^{x}t^{2n}\,dt = \frac{x^{2n + 1}}{2n + 1} \leq \frac{1}{2n + 1}$$ and taking limits as $n \to \infty$ and using Squeeze theorem we get $$\lim_{n \to \infty}R_{n}(x) = 0$$ It follows that $$\lim_{n \to \infty}(-1)^{n}R_{n}(x) = 0$$ and as noted earlier this equation holds for all values of $x$ with $|x|\leq 1$. Now taking limits as $n \to \infty$ on both sides of equation $(4)$ we get $$\arctan x = x - \frac{x^{3}}{3} + \cdots = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{x^{2n - 1}}{2n - 1}$$ for all $x$ with $|x| \leq 1$.
