Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.

My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x)^2}=\sum_{n=1}^\infty (-1)^{n+1}nx^{n-1}=\sum_{n=0}^\infty(-1)^n(n+1)x^n$$
From these,
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{4}+\sum_{n=0}^\infty \frac{(-1)^nx^n}{4}+\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)x^n}{2} \\ =\sum_{n=0}^\infty\frac{x^n+(-1)^nx^n+(-1)^{n+1}(n+1)x^n}{4} \\ =\sum_{n=0}^\infty\frac{(1+(-1)^n+(-1)^{n+1}(n+1))x^n}{4}$$
$f^{(2020)}(0)=(2020)!\times a_{2020}$ where $f(x)=\sum a_nx^n$
In this case, $a_n = 1+(-1)^n+(-1)^{n+1}(n+1), a_{2020}=-2019$ but there is no same answer.
Where did I make a mistake?

We have
$$f(x) = \frac14 \cdot \frac{1}{1-x}
+ \frac14 \cdot \frac{1}{1 + x}
+ \frac12 \cdot \left(\frac{1}{1 + x}\right)'.$$
By observing the first several derivatives to see the pattern:
$$\left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2}, 
\left(\frac{1}{1-x}\right)'' = \frac{2}{(1-x)^3},
\left(\frac{1}{1-x}\right)''' = \frac{3}{(1-x)^4}, \cdots$$
$$\left(\frac{1}{1+x}\right)' = \frac{-1}{(1+x)^2}, 
\left(\frac{1}{1+x}\right)'' = \frac{2}{(1+x)^3},
\left(\frac{1}{1+x}\right)''' = \frac{-3}{(1+x)^4}, \cdots$$
we have
$$\left(\frac{1}{1-x}\right)^{(n)}
= \frac{n!}{(1-x)^{n+1}}$$
and
$$\left(\frac{1}{1+x}\right)^{(n)}
= \frac{(-1)^n n!}{(1+x)^{n+1}}.$$
Thus, we have
$$f^{(n)}(x) = \frac14 \cdot \frac{n!}{(1-x)^{n+1}}
+ \frac14 \cdot \frac{(-1)^n n!}{(1+x)^{n+1}} + \frac12 \cdot \frac{(-1)^{n+1} (n+1)!}{(1+x)^{n+2}} $$
and
$f^{(2020)}(0) = -1010 \cdot 2020!$.
