Complicated partial fraction expansion I'm reading the book generatingfunctionology by Herbert Wilf and I came across a partial fraction expansion on page 20 that I cannot understand. The derivation is as follows:
$$
\frac{1}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \frac{\alpha_j}{1-jx}
$$
The book says to fix $r, 1 \leq r \leq k$, and multiply both sides by $1-rx$. Doing so, I get:
$$
\frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)} = \frac{\alpha_1(1-rx)}{1-x} + ... + \frac{\alpha_{r-1}(1-rx)}{1-(r-1)x} + \alpha_r + \frac{\alpha_{r+1}(1-rx)}{1-(r+1)x} + ... + \frac{a_k(1-rx)}{1-kx}
$$
Contrarily, the book has:
$$
\alpha_r = \frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)}
$$
I don't understand how the other other fractions on the right side of my result cancel out to $0$. I tried with a small example where $k=3$ and I couldn't isolate $\alpha_2$ nicely after multiplying both sides by $1-2x$. Any pointers on this would be greatly appreciated.
After this, the book goes on by letting $x=1/r$, resulting in the following:
$$
\begin{aligned}
\alpha_r &= \frac{1}{(1-1/r)(1-2/r)...(1-(r-1)/r)(1-(r+1)/r)...(1-k/r)} \\
 &= (-k)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!} && (1 \leq r \leq k)
\end{aligned}
$$
I also can't figure how this is derived (I suspect it's using an identity that I'm not aware of.) Any help would be much appreciated. Thanks a lot!
 A: It appears that the $\alpha_j$ coefficients are being computed by the standard Heaviside method (sometimes called the "cover up" method).
Heaviside's Method for partial fraction expansion
A: My edition of the book doesn’t have
$$\alpha_r = \frac{1}{(1-x)\ldots(1-(r-1)x)(1-(r+1)x)\ldots(1-kx)}\;,$$
which in any case makes no sense, since $\alpha_r$ is a constant and the righthand side is not. However, after letting $x=\frac1r$ it does show
$$\begin{align*}
\alpha_r&=\frac1{(1-1/r)(1-2/r)\cdots(1-(r-1)/r)(1-(r+1)/r)\cdots(1-k/r)}\\
&=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;.
\end{align*}$$
The last step is obtained by first multiplying by $r^{k-1}$ to get
$$\alpha_r=\frac{r^{k-1}}{\underbrace{(r-1)(r-2)\cdot(r-(r-1))}_{(r-1)!}\underbrace{(r-(r+1))\cdots(r-k)}_{(-1)(-2)\cdots(-(k-r))}}$$
and then rewriting this as
$$\alpha_r=\frac{r^{k-1}}{(r-1)!(-1)^{k-r}(k-r)!}=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;.$$
A: To find the coefficients of the expansion, you can follow the usual "cover" up rule which is essentially the results you have shown albeit buried in indices and what not:  $$
\frac{1}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \frac{\alpha_j}{1-jx}
$$
say we would want to find an arbitrary $\alpha_j$ for some value of $j \in \{1,2,...,k\}$. We call this $r$ to emphasise, and to avid confusion with the summation index. So we multiply both sides by $1-rx$
$$
\frac{1-rx}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \alpha_j\frac{1-rx}{1-jx}
$$
Since $r$ will coincide with one of the $j$'s we can "pull this out" from the right hand side
$$
RHS = \alpha_r + \sum_{\underset{j\ne r}{j=1}}^{k} \alpha_j\frac{1-rx}{1-jx}
$$
And for the left hand side, the r will "cancel" on one of the denominator factors:
$$
LHS = \frac{1}{\underset{\text{without the $1-rx$}}{(1-x)(1-2x)...(1-kx)}}
$$
The expansion holds for all values of x so we can set $x=1/r$, and simplify both sides to get the result. I'll stop here because I think doing it is fun :)
