<s>[INST] Is $(-1)^n$ needed in the power series of this function? To represent this function $$f(x)=\frac{2 x + 3}{x^{2} + 3 x + 2}$$
as the sum of a power series. 
My answer is $$\sum^{\infty}_{n=0}(-1)^{n}\left(\frac{1}{2^{n+1}}+1\right)x^{n}$$
$R=1$ and $I =(-1,1)$.
But the solution manual from Chegg is $$\sum^{\infty}_{n=0}\left(\frac{1}{2^{n+1}}+1\right)x^{n}$$ and $R = 2$, $I=(-2,2)$. Which I think it is wrong. 
But just want to double check with the people here. 
 [/INST] One has, by a partial fraction decomposition,
$$
\frac{2 x + 3}{x^{2} + 3 x + 2}=\frac{1}{x+1}+\frac{1}{x+2}
$$ then
$$
\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^n x^n,\qquad |x|<1,
$$
$$
\frac{1}{2+x}=\frac{1}{2}\cdot\frac{1}{1+\frac{x}2}=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}} x^n,\qquad |x|<2,
$$ giving

$$
\frac{2 x + 3}{x^{2} + 3 x + 2}=\sum_{n=0}^\infty (-1)^n\left(1+\frac{1}{2^{n+1}}\right) x^n,\qquad |x|<1.
$$

</s>