Answer
$\Sigma_{k=0}^{\infty} (-1) (x+1)^k $
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\dfrac{1}{x}$
Differentiate
$f'(x)=\dfrac{-1}{x^2}$
$f''(x)=\dfrac{2}{x^3}$
$f'''(x)=\dfrac{-6}{x^4}$
Now $f(-1)=-1$ , $f'(-1)=-1$ , $f''(-1)=-2$, $f'''(-1)=-6$
Hence the Maclaurin Series for the function is
$\displaystyle -1-(x+1)-(x+1^2- .......- (x+1)^n =\Sigma_{k=0}^{\infty} (-1) (x+1)^k $
