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] $ ?

A fast way is by exploiting $\arctan'(x)=\dfrac1{1+x^2}$.
The Taylor series of that derivative is easily established to be the sum of a geometric series of common ratio $-x^2$,
$$\sum_{k=0}^\infty(-x^2)^k=\frac1{1-(-x^2)},$$ which only converges for $x^2<1$.
Then integrating term-wise,
$$\arctan(x)=\int_0^x\frac{dx}{1+x^2}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}.$$

Alternatively, assume that you know
$$\ln(1+z)=-\sum_{k=1}^\infty\frac{(-z)^k}{k}.$$
Then with $z=ix$,
$$\ln(1+ix)=\ln\left(\sqrt{1+x^2}\right)+i\arctan(x)=-\sum_{k=1}^\infty\frac{(-ix)^k}{k}.$$
The imaginary part of this identity gives
$$\arctan(x)=\sum_{odd\ k}(-1)^{(k-1)/2}\frac{x^k}k$$
while the real part gives the extra
$$\ln\left(\sqrt{1+x^2}\right)=\sum_{even\ k>0}(-1)^{k/2-1}\frac{x^k}k.$$
