<s>[INST] 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?
 [/INST] A bit tricky
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$ Integrate $$g(x)=\int f(x)\,dx=\frac{1}{4} \left(\frac{2}{x+1}-\log (1-x)+\log (x+1)\right)$$
$$g(x)=\frac 12 +\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4 n} x^n$$ Differentitate
$$f(x)=g'(x)=\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4} x^{n-1}$$
Now, to make $x^{n-1}=x^{2020}$, $n=2021$ and then the coefficient of $x^{2020}$ is $-1010$.
Just multiply by $2020!$
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