Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\left(1+\frac{x^n}{3^n}\right)$ as a function of the real parameter $x$.
Explanation of Method
With $a_n=(x/3)^n$, absolute convergence of the product is governed by the geometric series $\sum|x/3|^n$, giving the threshold $|x|<3$. The three boundary/exterior cases ($x=3$, $x=-3$, $|x|>3$) each need separate elementary treatment since the general series criterion breaks down when $a_n\not\to0$.
- Apply the geometric-series convergence criterion to $a_n=(x/3)^n$.
- Check the boundary cases $x=\pm3$ directly.
- Confirm divergence for $|x|>3$, where terms no longer approach $1$.
Step 1 · The geometric threshold
Since $a_n=(x/3)^n$, $\sum|a_n|=\sum|x/3|^n$ is geometric and converges iff $|x/3|<1$, i.e. $|x|<3$. For $|x|<3$ this gives absolute (hence unconditional) convergence of the product to a positive finite limit, with $$ \sum_{n=1}^\infty\frac{x^n}{3^n}=\frac{x}{3-x}. $$
Step 2 · Boundary cases
At $x=3$: every factor is $1+1=2$, so $P_N=2^N\to\infty$ — diverges. At $x=-3$: the very first factor is $1+(-1)=0$ (since $(-3)^1/3^1=-1$), so every partial product from $N=1$ onward is exactly $0$.
Step 3 · Exterior case $|x|>3$
Here $|x/3|^n\to\infty$, so the factors $1+(x/3)^n$ do not tend to $1$ — the necessary condition for product convergence fails, and $P_N$ grows without bound (numerically confirmed: at $x=4$, $P_3\approx21.5$, growing rapidly).
$$ \boxed{\prod_{n=1}^{\infty}\left(1+\frac{x^n}{3^n}\right)\ \begin{cases}\text{converges to a positive finite limit} & |x|<3\\ =0 & x=-3\\ \text{diverges to }+\infty & x\ge3\text{ or }x<-3.\end{cases}} $$
Summary Table
| Range of $x$ | Behavior |
|---|---|
| $|x|<3$ | converges (absolutely) to a positive finite limit |
| $x=3$ | diverges: $P_N=2^N\to\infty$ |
| $x=-3$ | $P_N=0$ for all $N\ge1$ |
| $|x|>3$ or $x<-3$ | diverges |
Additional Notes
- The convergence interval $(-3,3)$ is an infinite-product analogue of a power series' radius of convergence — here the "radius" $3$ comes directly from the base of the exponential $x^n/3^n$, exactly as it would for the geometric series $\sum(x/3)^n$ itself.
- The degenerate case $x=-3$ (product identically $0$ from the first factor onward) is a useful reminder that a single vanishing factor kills the entire infinite product, regardless of the behavior of all later factors — the same phenomenon seen at $\alpha=\pm1$ in Problem 378.
- For $0\le x<3$ this product is a special case of the $q$-Pochhammer-type products studied in Problem 378 (there with base $\alpha$ directly; here with $\alpha=x/3$), reinforcing that both problems reduce to the same underlying geometric-series convergence mechanism.