## Thomas' Calculus 13th Edition

$1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...$ or, $\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$
Differentiate the given function $f(x)$ as follows: $f'(x)=\dfrac{-2}{x^3} \implies f'(1)=1$; $f''(x)=\dfrac{-6}{x^4} \implies f''(1)=-2\\ f'''(x)=\dfrac{-24}{x^5}\implies f'''(1)=-24$ The Taylor series at $x=1$ can be written as follows: $f(x)=f(1)+f'(-1) (x-1)+\dfrac{f''(1) }{2!} (x-1)^2+\dfrac{f''(1)}{3!}(x-1)^3 +\dfrac{f''(1) }{4!}(x-1)^4+.....\\=1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...=\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$