Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 51

$$\dfrac{-1}{(1+x)^2} ; |x| \lt 1 $$

The Taylor series for $\dfrac{1}{1-x} $ can be defined as: $\dfrac{1}{1-x}= 1+x+x^2+......+x^n+.....; |x| \lt 1$ Consider the given series: $\\=-1+2x-3x^2+4x^3-.....\\=\dfrac{d}{dx} [1-x+x^2-x^3+x^4-......]\\=\dfrac{d}{dx} (1/1+x) \\=\dfrac{-1}{(1+x)^2} ; |x| \lt 1 $