What's the Maclaurin Series of $f(x)=\frac{1}{(1-x)^2}$? This function seemed to be pretty much straight forward, but my solution is incorrect.
I have two questions:
1. Where did I make a mistake?
2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?
$$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$
$$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$ 
$$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$ 
$$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$ 
$$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$
The series is then
$$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$
and when simplified is 
$$1+x+x^2+x^3+x^4$$
But the correct answer is 
$$1+2x+3x^2+4x^3+5x^4 $$
 A: Your answer is very good, except the factorials in the denominators are off by one.  For example it should be $\frac{f''(0)}{2!}x^2$.
Shortcut:  $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ is a geometric series.  Take the derivative of both sides to get $\frac{1}{(1-x)^2}=1+2x+3x^2+\cdots$.
A: There’s a method to find this out that doesn’t use Calculus at all, and is comprehensible to a high-school student. Just do long division on the problem as you were taught in h.s., but writing the denominator as $1-2x+x^2$, and getting tems in the quotient of ever higher degree. When you get to the $4x^3$ term, you realize that something regular is going on.
A: Recall that
$$f(x) = f(0) + \dfrac{f'(0)}{1!} x + \dfrac{f''(0)}{2!}x^2 + \cdots$$
Your derivative computation is correct, i.e., $f^{(n)}(0) = (n+1)!$. But you have them matched to the wrong denominators. Hence, you will get
$$\dfrac1{(1-x)^2} = 1 + \dfrac{2!}{1!}x + \dfrac{3!}{2!}x^2 + \dfrac{4!}{3!}x^3 + \cdots = 1 + 2x + 3x^2 + 4x^3 + \cdots$$
A: As far as shortcuts go, there are two you might find interesting.
You've probably seen the geometric series several times:
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$
Now the first idea is that your function $f(x)$ is the square of this one, so:
$$ f(x) = (1 + x + x^2 + x^3 + \ldots )^2 $$
If the indicated multiplication is carried out, one gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
The second idea is to differentiate the geometric series formula:
$$ \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2} = f(x) $$
Thus the term-by-term differentiation of the geometric series also gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
A: Do not forget the binomial theorem approach
$$ (1-x)^{-2}=\sum_{k=0}^{\infty} {-2\choose k}(-x)^k = \dots\,, $$
where $$ {n\choose k}=\frac{n!}{(n-k)!k!}. $$
