Answer
$-1-2x-\dfrac{5x^2}{2}+....$
Work Step by Step
The Maclaurin series for $\dfrac{2}{1-x}$ can be written as:
$2(1+x+x^2+....+x^n+...)=(2) \Sigma_{n=0}^{\infty} x^n$
Now, $\cos x -\dfrac{2}{1-x}=1 -\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+......-[2(1+x+x^2+....+x^n+...)] \\ \implies \cos x -\dfrac{2}{1-x}=[\Sigma_{n=0}^{\infty}\dfrac{ (-1)^n
x^{2n}}{(2n)!}]-[(2) \Sigma_{n=0}^{\infty} x^n] =-1-2x-\dfrac{5x^2}{2}+....$
