Answer
$\Sigma_{n=0}^{\infty} (-1)^{n} (x-1)^n$
Work Step by Step
Here, we have: $f(x)=\dfrac{1}{x}\implies f(1)=1\\ f'(x)=-\dfrac{1}{(x^2} \implies f'(1)=-1\\f''(x)=\dfrac{2}{x^3} \implies f''(1)=2$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f′(a)}{1!}(x−a)+\dfrac{f′′(a)}{2!}(x−a)^2+......+\dfrac{f^n(a)}{n!}(x−a)^n$
Hence, the required Taylor series is:
$T(x)=1-(x-1)+\dfrac{2}{2!}(x-1)^2+.......=\Sigma_{n=0}^{\infty} (-1)^{n} (x-1)^n$
