Question

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 $\Pi_{i=0}^n (1-a_i)$?

Answer 1

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.

Answer 2

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.
\[ (5) \quad (4+1)\quad (3+2)\quad(6+(-1))\quad (5+0) \quad (4+1+0)\quad (3+2+0)\quad(6+0+(-1))\dots\]

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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