TITLE: Calculating value of $1000^{th}$ derivative at $0$.
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: 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
$$f^{(n)}(0)=n!\left(2-\frac3{2^{n+1}}\right)\;.$$