Answer
$f(x)=1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...=\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$
Work Step by Step
We have $f'(x)=\dfrac{-2}{x^3} \implies f'(1)=1$;
$f''(x)=\dfrac{-6}{x^4} \implies f''(1)=-2\\ f'''(x)=\dfrac{-24}{x^5}\implies f'''(1)=-24$ and $f^{4}(1)=-120$ and so on.
Therefore, the Taylor's series at $x=1$ is as follows:
$f(x)=f(1)+f'(-1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f''(1)(x-1)^3 }{3!}+\dfrac{f''(1)(x-1)^4 }{4!}+.....so \space on$
or, $f(x)=1-2(x-1)+3(x-1)^2-4(x-1)^3+5(x-1)^4+...=\Sigma_{n=0}^{\infty} (-1)^n (n+1)(x-1)^n$
