# Stirling Numbers and partial fraction decomposition

I wanted to ask how can I break into partial fraction this expression: $$\frac {1}{(1-x)(1-2x)...(1-kx)}$$ I tried saying for $x=\frac{1}{i}$ we get some expression for $A_i$ but got stuck.

Thank you very much!

$$f(x)=\frac{1}{\prod_{k=1}^{n}(1-kx)}=\sum_{k=1}^{n}\frac{A_k}{x-\frac{1}{k}},$$

$$A_k = \text{Res}\left(f(x),x=\frac{1}{k}\right) = \frac{-1}{k\prod_{\substack{1\leq j\leq n\\ j\neq k}}\left(1-\frac{j}{k}\right)}$$

$$A_k = \frac{-k^{n-1}}{k\prod_{\substack{1\leq j\leq n\\ j\neq k}}(k-j)}=\color{red}{\frac{k^{n-1}(-1)^{n-k+1}}{k!(n-k)!}}.$$

• You might want to check your arithmetic on this one. – Marko Riedel Jan 21 '17 at 21:47
• While I might have gone too far in preparing for residue extraction you need the term $\frac{1}{x-1/k}$ in your product for it to be legal to substitute into the rest for the residue. This explains the missing $-k.$ – Marko Riedel Jan 21 '17 at 22:37
• @MarkoRiedel: all right, many thanks. And (+1) back. – Jack D'Aurizio Jan 21 '17 at 22:38
• Thanks to both of you! – Omer.S Jan 23 '17 at 19:27

As the poles are all simple we can use partial fractions by residues. With

$$f(z) = \prod_{k=1}^n \frac{1}{1-kz}$$

we have

$$f(z) = \sum_{k=1}^n \frac{1}{z-1/k} \mathrm{Res}_{z=1/k} f(z).$$

To extract these residues we write

$$f(z) = \frac{1}{n!} \prod_{k=1}^n \frac{1}{1/k-z} = \frac{(-1)^n}{n!} \prod_{k=1}^n \frac{1}{z-1/k}$$

We then get for the residue

$$\mathrm{Res}_{z=1/k} f(z) = \frac{(-1)^n}{n!} \prod_{j=1, j\ne k}^n \frac{1}{1/k-1/j} \\ = \frac{(-1)^n}{n!} \prod_{j=1, j\ne k}^n \frac{kj}{j-k} = \frac{(-1)^n}{n!} k^{n-1} \prod_{j=1, j\ne k}^n \frac{j}{j-k} \\ = \frac{(-1)^n}{n!} k^{n-1} \frac{n!}{k} \frac{(-1)^{k-1}}{(k-1)! (n-k)!} = (-1)^{n+k-1} k^{n-1} \frac{1}{k! (n-k)!}.$$