Answer
$\Sigma_{k=0}^n-(x+1)^k$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_x(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
Given function, $f(x)=\dfrac{1}{x}$
Differentiate w.r.t x
$f'(x)=-\dfrac{1}{x^2}$
$f''(x)=\dfrac{2}{x^3}$
$f'''(x)=-\dfrac{6}{x^4}$
$f''''(x)=\dfrac{24}{x^5}$
Now $f(-1)=1$, $f'(-1)=-1$, $f''(-1)=-2$, $f'''(-1)=-6$ and $f''''(-1)=-24$
Hence the Maclaurin Series for the function is
$\displaystyle -1-(x+1)-(x+1)^2- .......-(x+10)^n=\Sigma_{k=0}^n-(x+1)^k$
