Proof of the pentagonal number theorem

Jordan Bell
April 3, 2014

Let A0=k=1(1-zk). We will use the identity


which is straightforward to prove by induction. We apply the identity with ak=zk and N=, which gives

A0 = 1-z-k=2zk(1-z)(1-zk-1)
= 1-z-k=0zk+2(1-z)(1-zk+1).

For n1 let An=k=0znk(1-zn)(1-zn+k). We have A0=1-z-z2A1, and for n1 we have

An = 1-zn+k=1znk(1-zn)(1-zn+k)
= 1-zn+k=1znk(1-zn+1)(1-zn+k)
= 1-zn+zn(1-zn+1)+k=2znk(1-zn+1)(1-zn+k)
= 1-z2n+1+k=0zn(k+2)(1-zn+1)(1-zn+k+2)
= 1-z2n+1-k=0zn(k+2)+n+k+2(1-zn+1)(1-zn+k+1)
= 1-z2n+1-z3n+2k=0z(n+1)k(1-zn+1)(1-zn+k+1)
= 1-z2n+1-z3n+2An+1.

Therefore An=1-z2n+1-z3n+2An+1 for all n0.

We then check by induction that for all M

A0 = 1-z+n=1M(-1)n(zn(3n+1)/2-z(n+1)(3n+2)/2)

and taking M= gives the pentagonal number theorem.