Maclaurin series for $\dfrac{\cos(x)-1}{x^2}$

Question

The solution for this is $$ -\dfrac{1}{x} + \dfrac{x^2}{4!} - \dfrac{x^2}{6!} + \ldots \;, $$ but I'm not sure how to derive this Maclaurin series from $\cos(x)$. The solution just divided each term in the Maclaurin series for $\cos(x)$ by $x^2$, and then subtracted out the beginning term in $\cos(x)$ (which is $1$ in $1+x^2/2!-\ldots$). Dividing by $x^2$ makes sense, but how come only the first $1$ is subtracted? Shouldn't a $1$ be subtracted for each term and then the term be divided by $x^2$?

Answer 1

$$\cos x=\sum^\infty_{k=0}\dfrac{(-1)^kx^{2k}}{(2k)!}\\ \implies \cos x-1=\sum^\infty_{k=1}\dfrac{(-1)^kx^{2k}}{(2k)!}\\ \implies \dfrac{\cos x-1}{x^2}=\sum^\infty_{k=1}\dfrac{(-1)^kx^{2(k-1)}}{(2k)!}$$ Expand the series.

Answer 2

Since $$ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb $$ you have $$ \cos x - 1 = -\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb $$ and so $$ \frac{\cos x - 1}{x^2} = -\frac{1}{2!}+\frac{x^2}{4!}-\frac{x^4}{6!}+\dotsb $$