Proof of the pentagonal number theorem

Jordan Bell

April 3, 2014

Let $A_{0}=\prod_{k=1}^{\infty}(1-z^{k})$ . We will use the identity

\prod_{k=1}^{N}(1-a_{k})=1-a_{1}-\sum_{k=2}^{N}a_{k}(1-a_{1})\cdots(1-a_{k-1}),

which is straightforward to prove by induction. We apply the identity with $a_{k}=z^{k}$ and $N=\infty$ , which gives

	$\displaystyle A_{0}$	$\displaystyle=$	$\displaystyle 1-z-\sum_{k=2}^{\infty}z^{k}(1-z)\cdots(1-z^{k-1})$
		$\displaystyle=$	$\displaystyle 1-z-\sum_{k=0}^{\infty}z^{k+2}(1-z)\cdots(1-z^{k+1}).$

For $n\geq 1$ let $A_{n}=\sum_{k=0}^{\infty}z^{nk}(1-z^{n})\cdots(1-z^{n+k})$ . We have $A_{0}=1-z-z^{2}A_{1}$ , and for $n\geq 1$ we have

$\displaystyle A_{n}$	$\displaystyle=$	$\displaystyle 1-z^{n}+\sum_{k=1}^{\infty}z^{nk}(1-z^{n})\cdots(1-z^{n+k})$
	$\displaystyle=$	$\displaystyle 1-z^{n}+\sum_{k=1}^{\infty}z^{nk}(1-z^{n+1})\cdots(1-z^{n+k})$
		$\displaystyle-\sum_{k=1}^{\infty}z^{n(k+1)}(1-z^{n+1})\cdots(1-z^{n+k})$
	$\displaystyle=$	$\displaystyle 1-z^{n}+z^{n}(1-z^{n+1})+\sum_{k=2}^{\infty}z^{nk}(1-z^{n+1})% \cdots(1-z^{n+k})$
		$\displaystyle-\sum_{k=1}^{\infty}z^{n(k+1)}(1-z^{n+1})\cdots(1-z^{n+k})$
	$\displaystyle=$	$\displaystyle 1-z^{2n+1}+\sum_{k=0}^{\infty}z^{n(k+2)}(1-z^{n+1})\cdots(1-z^{n% +k+2})$
		$\displaystyle-\sum_{k=0}^{\infty}z^{n(k+2)}(1-z^{n+1})\cdots(1-z^{n+k+1})$
	$\displaystyle=$	$\displaystyle 1-z^{2n+1}-\sum_{k=0}^{\infty}z^{n(k+2)+n+k+2}(1-z^{n+1})\cdots(% 1-z^{n+k+1})$
	$\displaystyle=$	$\displaystyle 1-z^{2n+1}-z^{3n+2}\sum_{k=0}^{\infty}z^{(n+1)k}(1-z^{n+1})% \cdots(1-z^{n+k+1})$
	$\displaystyle=$	$\displaystyle 1-z^{2n+1}-z^{3n+2}A_{n+1}.$

Therefore $A_{n}=1-z^{2n+1}-z^{3n+2}A_{n+1}$ for all $n\geq 0$ .

We then check by induction that for all $M$

	$\displaystyle A_{0}$	$\displaystyle=$	$\displaystyle 1-z+\sum_{n=1}^{M}(-1)^{n}\Big{(}z^{n(3n+1)/2}-z^{(n+1)(3n+2)/2}% \Big{)}$
			$\displaystyle+(-1)^{M+1}z^{(M+1)(3M+2)/2}A_{M+1},$

and taking $M=\infty$ gives the pentagonal number theorem.