# Product of one minus geometric progression

Consider the geometric sequence $a_n:=ar^{n-1}$, $n=0,1,\ldots$.

We know that the product $\Pi_{i=0}^n a_i=(\sqrt{a_1\cdot a_{n+1}})^{n+1}$.

Is there a formula for $\Pi_{i=0}^n (1-a_i)$?

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The product equation isn't quite right. For $n=1$, for example, it equates $(a/r)a$ with $a(ar)$. The expression inside the square root, I believe, should be $a_0 a_n$. – Barry Cipra Jan 20 '12 at 16:27

In a standard $q$-notation, the product of your interest is $$(z;q) _ n:=\prod_{j=0}^{n-1}(1-zq^j).$$ The formula known as the $q$-binomial theorem expresses this product as sum: $$(z;q) _ n=\sum _{k=0}^n {\genfrac{[}{]}{0pt}{}{m}{n}} _q (-z)^kq^{k(k-1)/2},$$ where the $q$-binomial coefficients are defined, e.g., in this question.

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If $a = -1$ you get $\displaystyle \left(1 + \frac{1}{r}\right)\cdot 0 \cdot \prod_{i=k}^n (1 + r^k)= 0$. The last factor is a generating function (free book inside!) counting the number of partitions of $k$ into distinct parts with parts at length at most $n$.

Let's look at the partitions of $5$. $(5) \quad (4+1)\quad (3+2)\quad (3+1+1)\quad (2+2+1)\quad (2+1+1+1)\quad (1+1+1+1+1)$ This would appear as the $r^5$ term in that series. However, many of these do not have distinct parts. So we rule out the last four. $(5) \quad (4+1)\quad (3+2)$ However, with your two additional factors, you're allowing upto one $0$ and one $-1$ in your partition.
$(5) \quad (4+1)\quad (3+2)\quad(6+(-1))\quad (5+0) \quad (4+1+0)\quad (3+2+0)\quad(6+0+(-1))\dots$

Your sequence begins $\frac{a}{r}, a, ar, ar^2, ar^3, \dots$. The coefficient of $\displaystyle \prod_{i=0}^n (1 + br^{i-1})$ gives a weighted sum of partitions into distinct parts. Let's show how that count works. (NOTE: my $b$ is your $-a$).

• 5 has weight $b$
• (4 + 1) and (3+2) have weight $b^2$
• (3+2+0) has weight $b^3$ and so on.

In light of this, I doubt there is a closed formula for this infinite product beyond this or similar interpretations as counting partitions, (see Wadim's answer) but these are extremely interesting functions!

Two guys who study partitions for a living are Herbert Wilf and George Andrews

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Unfortunately, Herbert Wilf is no longer studying (as of Jan. 7, or so I understand). Gerhard "Ask Me About System Design" Paseman, 2012.01.20 – Gerhard Paseman Jan 20 '12 at 16:24
That is unfortunately correct. Fortunately, at least for the time being his excellent book "generatingfunctionology" can still be downloaded online: math.upenn.edu/~wilf/DownldGF.html – Michael Lugo Jan 20 '12 at 23:48