Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$

Question

Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.

Answer 1

\begin{align} & \lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})] \\ &= \lim_{n\to\infty}\frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\ &= \cdots \cdots \cdots \\ &= \lim_{n\to\infty}\frac{(1-x^{2^n})(1+x^{2^n})}{1-x} \\ &= \lim_{n\to\infty}\frac{1-x^{2^{n+1}}}{1-x} \\ &= \frac{1}{1-x}\lim_{n\to\infty}{(1-x^{2^{n+1}})} \\ &= \frac{1}{1-x}\cdot 1 \\ &= \frac{1}{1-x} . \end{align}

Answer 2

The product expands out as $$ (1 + x)(1 + x^2)(1 + x^4)\ldots(1 + x^{2^n}) = \sum_{k=0}^{2^{n+1} -1} x^k = \frac{1 - x^{2^{n+1}}}{1 - x}. $$

Since $|x| < 1$, this converges to $\frac{1}{1 - x}$ as $n \to \infty$.