Taylor Series of $\sin x/(1-x)$ Ιs there any fast way to calculate the first four non-zero terms a Taylor Series $\dfrac {\sin x}{1-x}$ at $x=0$ without making big derivatives calculations?
I know that $$\sin x = x- \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} \dots$$
and $$\frac{1}{1-x} = 1 + x + x^2 + x^3 \dots$$
Can we combine them together?

Maybe
$$(1-x)(a_0+a_1x+a_2x^2+a_3x^3\cdots)=x-\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}\cdots$$
$$a_0=0\\a_1-a_0=1\\a_2-a_1=0\\a_3-a_2=-\frac1{3!}\\a_4-a_3=0\\a_5-a_4=\frac1{5!}\\\cdots$$
so that
$$a_{2k+2}=a_{2k+1}=\sum_{i=0}^k\frac{(-1)^i}{(2i+1)!}.$$
The coefficients are the successive Taylor approximations of $\sin(1)$.
