Prove that product from n=0 to infinity
Prove that $\displaystyle\prod_{n=0}^{\infty}\left(1+x^{2^n}\right)=\frac1{1-x}$ for $|x|<1$.
Explanation of Method
Repeatedly applying $(1-a)(1+a)=1-a^2$ shows that multiplying $1-x$ by the first $N+1$ factors collapses everything to $1-x^{2^{N+1}}$; dividing back out and letting $N\to\infty$ (using $x^{2^{N+1}}\to0$ for $|x|<1$) gives the closed form.
- Establish the telescoping identity $(1-x)\prod_{n=0}^N(1+x^{2^n})=1-x^{2^{N+1}}$ by induction.
- Solve for the partial product $P_N$.
- Take the limit as $N\to\infty$.
Step 1 · The telescoping identity
By induction on $N$: base case $(1-x)(1+x)=1-x^2$. Inductive step: if $(1-x)\prod_{n=0}^N(1+x^{2^n})=1-x^{2^{N+1}}$, multiplying both sides by $(1+x^{2^{N+1}})$ gives $(1-x^{2^{N+1}})(1+x^{2^{N+1}})=1-x^{2^{N+2}}$, extending the identity to $N+1$. Hence for all $N\ge0$: $$ (1-x)\prod_{n=0}^N\left(1+x^{2^n}\right)=1-x^{2^{N+1}}. $$
Step 2 · Solve for the partial product
$$ P_N=\prod_{n=0}^N\left(1+x^{2^n}\right)=\frac{1-x^{2^{N+1}}}{1-x}. $$
Step 3 · Take the limit
For $|x|<1$, $x^{2^{N+1}}\to0$ as $N\to\infty$, so $$ \boxed{\prod_{n=0}^{\infty}\left(1+x^{2^n}\right)=\lim_{N\to\infty}\frac{1-x^{2^{N+1}}}{1-x}=\frac1{1-x}} $$ (Numerical check at $x=0.3$: $P_{20}\approx1.428571$ vs. $1/(1-0.3)=1.428571$ — matching exactly.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{1-x^{2^{N+1}}}{1-x}$ |
| Limit | $\dfrac1{1-x}$ |
| Domain | $|x|<1$ |
Additional Notes
- Correction to the stated index: as originally posed with the product starting at $n=1$, the identity $\prod_{n=1}^\infty(1+x^{2^n})=1/(1-x)$ is numerically false — direct computation at $x=0.3$ gives $P\approx1.098901$, matching $1/(1-x^2)\approx1.098901$, not $1/(1-x)\approx1.428571$. The telescoping identity $(1-x)\prod_{n=0}^N(1+x^{2^n})=1-x^{2^{N+1}}$ requires the $n=0$ factor $(1+x)$ to start the cancellation; omitting it leaves an uncancelled $(1+x)$ in the denominator, so $\prod_{n=1}^\infty(1+x^{2^n})=1/(1-x^2)$ instead. The classical result $1/(1-x)$ — which matches the stated closed form — requires starting the product at $n=0$, as used above.
- Expanding the product, each factor $(1+x^{2^n})$ contributes a binary digit: multiplying them all out generates exactly one term $x^k$ for every non-negative integer $k$, since every $k$ has a unique binary representation as a sum of distinct powers of $2$ — this is precisely why the product reproduces the full geometric series $\sum_{k\ge0}x^k=1/(1-x)$.
- This shares its telescoping engine with Problem 363 ($(1-a)(1+a)=1-a^2$ applied to $a=2^{-n}$-type terms) and Problem 365 (the same doubling structure via $\sinh(2\theta)=2\sinh\theta\cosh\theta$) — all three problems in this batch exploit a "square-and-halve" telescoping mechanism tied to powers of $2$.