Please help me to find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^2(x-1)}$ around $a=2$ First, sorry if my translations is bad. I need help for this exercise, more precisely , I need to know if my result which I've found is good.
The exercise: Find Taylor expansion (or approximation) for $f(x)=\frac{1}{x^2(x-1)}$ around point $2$. $a=2$ in Taylor series formula: $f(x)=\sum\limits_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$ 
I tried this: $f(x)=-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x-1}$ then I found $n$-th derivative of each function at point $a=2$:
$f_1(x)=-\frac{1}{x}; f_1^{(n)}(2)=(-1)^{n+1}\frac{n!}{2^{n+1}}$
$f_2(x)=-\frac{1}{x^2}; f_2^{(n)}(2)=(-1)^{n-1}\frac{(n+1)!}{2^{n+2}}$
$f_3(x)=\frac{1}{x-1}; f_3^{(n)}(2)=(-1)^n n!$
Then I wrote this: 
$$\begin{align}
f(x)&=\sum\limits_{n=0}^\infty(-1)^{n+1}\frac{(x-2)^n}{2^{n+1}}+\sum\limits_{n=0}^\infty(-1)^{n-1}\frac{(n+1)(x-2)^n}{2^{n+2}}+\sum\limits_{n=0}^\infty(-1)^n(x-2)^n\\
&=\sum\limits_{n=0}^\infty(-1)^{n+1}(x-2)^n\frac{2^{n+2}+n+3}{2^{n+2}}\end{align}$$
I hope someone could help me by telling me if result wich I've found is correct or not.
 A: The general procedure is correct.  And the details look pretty good too.  I could now check the details,  it would not be difficult. 
Instead, I will do the problem in a somewhat different way, which in general is more efficient.  It will turn out that a sign error crept into your calculation.  
We have 
$$f(x)=-\frac{1}{x}-\frac{1}{x^2} + \frac{1}{x-1}$$
and want to express $f(x)$ as a sum of powers of $x-2$.  It is useful, though not necessary, to let $y=x-2$.  Then $x=y+2$.  Substituting for $x$, we obtain
$$f(y+2)=g(y)=  -\frac{1}{y+2}-\frac{1}{(y+2)^2} + \frac{1}{y+1}.$$
We want to express $g(y)$ as a sum of powers of $y$.  Let's start with the easiest part, $\frac{1}{1+y}$.
It would be a shame to do a whole lot of differentiating when we already know the power series expansion of $1/(1+y)$.  Or at least we certainly know the power series expansion of $1/(1-z)$, and then we can put $y=-z$. Thus
$$\frac{1}{1+y}=1-y+y^2-y^3+\cdots=\sum_0^\infty (-1)^ny^n\qquad\qquad\text{(Term $1$)}$$
That was easy.  Let's go on to the next easiest term, $1/(y+2)$ (I know there should be a minus sign in front, will take care of it later). We have 
$$\frac{1}{2+y}=\frac{1/2}{1+y/2}.$$
It would be a shame not to use the fact that we know the power series expansion of $1/(1-z)$.  We get
$$\frac{1}{2+y}=\frac{1}{2}\sum_0^\infty (-1)^n\frac{1}{2^n}y^n=\sum_0^\infty \frac{(-1)^n}{2^{n+1}}y^n,$$ 
and therefore
$$-\frac{1}{2+y}=\sum_0^\infty \frac{(-1)^{n+1}}{2^{n+1}}y^n.\qquad\qquad\text{(Term $2$)}$$
Finally, we want the expansion of $1/(2+y)^2$ in powers of $y$.  We have just obtained the expansion of $1/(2+y)$.  Note that $1/(2+y)^2$ is (almost) the derivative of $1/(2+y)$.  To be precise, it is the negative of the derivative of $1/(2+y)$.  So let us differentiate the series we obtained for $1/(2+y)$ term by term. We find
$$-\frac{1}{(y+2)^2}=\sum_0^\infty \frac{(-1)^{n} n}{2^{n+1}}y^{n-1}=\sum_0^\infty \frac{(-1)^{n+1} (n+1)}{2^{n+2}}y^{n} \qquad\qquad\text{(Term $3$)}$$
Now it is just a matter of adding Terms $1$, $2$, and $3$ together. I get, replacing $y$ by $x-2$, the following:
$$\sum_0^\infty (-1)^n\left(1-\frac{n+3}{2^{n+2}}\right)(x-2)^n.$$
Comment: When we are calculating power series expansions, it is nice to avoid all those differentiations, by recycling standard expansions.    
