Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}(1-\alpha^n)$ as a function of the real parameter $\alpha$.
Explanation of Method
For $|\alpha|<1$, $\sum|\alpha|^n$ is a convergent geometric series, so the product converges absolutely to Euler's classical $q$-Pochhammer-type function $(\alpha;\alpha)_\infty$. The boundary values $\alpha=\pm1$ each produce an exactly-zero factor, and $|\alpha|>1$ makes the terms blow up rather than approach $1$, forcing divergence.
- Establish absolute convergence for $|\alpha|<1$ via the geometric series $\sum|\alpha|^n$.
- Identify the exactly-zero degenerate cases $\alpha=\pm1$.
- Show divergence for $|\alpha|>1$, where terms don't approach $1$.
Step 1 · Convergence for $|\alpha|<1$
Since $\sum_{n=1}^\infty|\alpha|^n=\dfrac{|\alpha|}{1-|\alpha|}$ converges, the product $\prod(1-\alpha^n)$ converges absolutely to a positive finite limit for $0\le\alpha<1$, and to a finite nonzero limit for $-1<\alpha<0$ (the sign pattern of individual factors alternates around $1$, but absolute convergence guarantees the limit exists and is nonzero). This is Euler's function, related to the $q$-Pochhammer symbol $(\alpha;\alpha)_\infty$, fundamental in partition theory and modular forms.
Step 2 · The degenerate boundary cases
At $\alpha=1$: the very first factor is $1-1=0$, so $\prod=0$. At $\alpha=-1$: the second factor is $1-(-1)^2=1-1=0$, so again $\prod=0$. In both cases a single vanishing factor forces the entire infinite product to be exactly $0$.
Step 3 · Divergence for $|\alpha|>1$
For $\alpha>1$, every factor $1-\alpha^n<0$ and $|1-\alpha^n|=\alpha^n-1\to\infty$; the partial products $P_N=(-1)^N\prod_{n=1}^N(\alpha^n-1)$ oscillate in sign with unboundedly growing magnitude, so the product diverges. For $\alpha<-1$, odd- and even-indexed factors have opposite, unboundedly growing behavior, again giving divergence.
$$ \boxed{\prod_{n=1}^{\infty}(1-\alpha^n)\ \begin{cases}\text{converges to a nonzero finite limit} & |\alpha|<1\\ =0 & \alpha=\pm1\\ \text{diverges} & |\alpha|>1.\end{cases}} $$ (Numerical check: at $\alpha=1/2$, $P_{20}\approx0.288788$, matching Euler's function value; at $\alpha=-1/2$, $P_{20}$ stabilizes to a finite nonzero value.)
Summary Table
| Range of $\alpha$ | Behavior |
|---|---|
| $0\le\alpha<1$ | converges (positive, absolute) |
| $-1<\alpha<0$ | converges (nonzero, absolute) |
| $\alpha=\pm1$ | exactly $0$ (a factor vanishes) |
| $|\alpha|>1$ | diverges |
Additional Notes
- No elementary closed form exists for $\prod(1-\alpha^n)$ at a general $0<\alpha<1$ — this is precisely Euler's function, whose reciprocal generates the partition-counting function via $1/\prod(1-\alpha^n)=\sum_kp(k)\alpha^k$; special values (e.g. via Euler's pentagonal number theorem, $\prod(1-\alpha^n)=\sum_k(-1)^k\alpha^{k(3k-1)/2}$) exist but the product itself is not "elementary" the way the telescoping products of Problems 360–366 were.
- This is structurally the direct generalization of Problem 373's geometric threshold: there the base was $x/3$ inside a $(1+a_n)$ product; here the base $\alpha$ appears directly, with the sign flipped to $(1-\alpha^n)$, but the same geometric-series convergence radius mechanism governs both.
- The degenerate zero at $\alpha=\pm1$ parallels Problem 373's $x=-3$ case: whenever a single factor of an infinite product vanishes identically, the entire product is $0$ regardless of how well-behaved every other factor is.