Calculating value of $1000^{th}$ derivative at $0$. I need to calculate value of $1000^{th}$ derivate of the following function at $0$:
$$
f(x) = \frac{x+1}{(x-1)(x-2)}
$$
I've done similar problems before (e.g. $f(x)= \dfrac{x}{e^{x}}$) but the approach I've used would not work in this case and I believe I should expand this function into a power series. Could you please give me any hints on how to do it?
 A: Hint: Note that $f(x)=\frac{3}{x-2}-\frac{2}{x-1}$ and if $g(x)=\frac{1}{x-a}$ then $g^n(x)=\frac{(-1)^n n! }{(x-a)^{n+1}}$ where $a$ is a constant and $g^n(x)$ is the $n$-th derivative of $g(x)$.
Here $$\begin{align} \frac{A}{x-2}+\frac{B}{x-1} &=\frac{x+1}{(x-2)(x-1)}\\ \implies A(x-1)+B(x-2) &= x+1 \\ \implies x(A+B)+(-A-2B) &= x+1\end{align}$$
From this we get 
$$\begin{align}A+B &=1 \\ -A-2B &=1 \end{align}$$
A: As suggested by Argha and David Mitra, begin with the decomposition into partial fractions:
$$
f(x)=\frac{3}{x-2}-\frac{2}{x-1}.
$$
Now if $g(x)=\frac{1}{x-a}$ for some constant $a$, then a proof by induction quickly shows that 
$$
g^{(n)}(x)=\frac{(-1)^nn!}{(x-a)^{n+1}}.
$$
It follows that 
$$
f^{(1000)}(x)=\frac{3\cdot 1000!}{(x-2)^{1001}}-\frac{2\cdot 1000!}{(x-1)^{1001}}.
$$
Now make $x=0$.
A: Starting with the partial fraction decomposition $$f(x)=\frac{3}{x-2}-\frac{2}{x-1}\;,$$ use the geometric series sum to write 
$$\begin{align*}
f(x)&=-\frac32\cdot\frac1{1-\frac{x}2}+2\frac1{1-x}\\\\
&=2\sum_{n\ge 0}x^n-\frac32\sum_{n\ge 0}\left(\frac{x}2\right)^n\\\\
&=\sum_{n\ge 0}\left(2-\frac32\left(\frac12\right)^n\right)x^n\\\\
&=\sum_{n\ge 0}\left(2-\frac3{2^{n+1}}\right)x^n\;.
\end{align*}$$
Then equate this with the Maclaurin series
$$f(x)=\sum_{n\ge 0}\frac{f^{(n)}(0)}{n!}x^n$$
to get $$\frac{f^{(n)}(0)}{n!}=2-\frac3{2^{n+1}}$$ and then
$$f^{(n)}(0)=n!\left(2-\frac3{2^{n+1}}\right)\;.$$
