I am trying exercises of Ch-14 partitions from Tom Apostol Introduction to ANT and unable to Solve (a) part of Question 5.
5. If $x\ne 1$ let $Q_0(x)=1$ and for $n\ge 1$ define
$$ Q_n(x) = \prod_{r=1}^n \frac{1-x^{2r}}{1-x^{2r-1}}. $$
(a) Derive the following finite identities of Shanks:
$$ \sum_{m=1}^{2n} x^{m(m-1)/2} = \sum_{s=0}^{n-1} \frac
{Q_n(x)}{Q_s(x)} x^{s(2n+1)}, $$
$$ \sum_{m=1}^{2n+1} x^{m(m-1)/2} = \sum_{s=0}^{n} \frac
{Q_n(x)}{Q_s(x)} x^{s(2n+1)}. $$
(b) Use Shank's identities to deduce Gauss' triangular-number
theorem:
$$ \sum_{m=1}^\infty x^{m(m-1)/2} = \prod_{n=1}^\infty
\frac{1-x^{2n}}{1-x^{2n-1}} \text{ for } |x|<1. $$
I have question in proof of (a) as I am unable to derive any of the identity. I tried by evaluating $Q_n(x) / Q_s(x) $ and if I multiply it by $ x^{s(2n+1) }$ I don't see anything to get LHS in (I).
So, kindly shed some light on how to prove it by proving any of the identity.
I shall be really thankful for your help.
Original image: https://i.sstatic.net/kUJbT.jpg
Best Answer
Main theme: We set for $n\geq 1$ \begin{align*} A_n(x)=\sum_{s=0}^{n-1}\frac{Q_n(x)}{Q_s(x)}x^{s(2n+1)}\qquad\qquad S_n(x)=\sum_{m=1}^{2n}x^{\frac{m(m-1)}{2}}\\ \end{align*} and find a representation of $A_n(x)$ via $A_{n-1}(x)$ by also separating terms which are summands from $S_n(x)$. \begin{align*} \color{blue}{A_n(x)}&=A_{n-1}(x)+x^{(n-1)(2n-1)}+x^{n(2n-1)}\\ &\,\,\color{blue}{=A_{n-1}(x)+S_{n}(x)-S_{n-1}(x)}\tag{1} \end{align*} We so derive by induction \begin{align*} A_n(x)-S_n(x)&=A_{n-1}(x)-S_{n-1}(x)=\cdots\\ &=A_{1}(x)-S_{1}(x)\\ &=\frac{1-x^2}{1-x}-(1+x)=0 \end{align*} and the claim follows.
Proof: The derivation of (1) is elementary but technically tricky.
Intermezzo:
We obtain for $s=0,\ldots,n-2$: \begin{align*} \color{blue}{\beta_{s+1,n}(x)}&=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{s+1}(x)}x^{(s+1)(2n-1)}\\ &=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}\,\frac{1-x^{2s+1}}{1-x^{2s+2}}x^{(s+1)(2n-1)}\\ &=\frac{1-x^{2s+1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{(s+1)(2n-1)}\\ &\,\,\color{blue}{=\alpha_{s,n}(x)}\tag{5} \end{align*}
We also have $\beta_{0,n}(x)=0$ and \begin{align*} \alpha_{n-1,n}(x)=\frac{1-x^{2n-1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{n-1}(x)}x^{n(2n-1)}=x^{n(2n-1)}\tag{6} \end{align*}
Comment:
In (7) we use the identity (4) and sum from $s=0$ to $s=n-1$. We set the lower index of the right-most sum to $s=1$ since $\beta_{0,n}(x)=0$.
In (8) we shift the index of the right-most sum by one to start with $s=1$.
In (9) we use the telescoping property $\alpha_{s,n}(x)-\beta_{s+1,n}(x)=0$ for $s=0,\ldots, n-2$.
In (10) we separate the term with $s=n-1$ from the sum and use the identity (6).
Note: This proof was given by D. Shanks in Two Theorems of Gauss in 1958.