Convergence of the infinite product $\prod _{n=1} ^\infty (1+x^n)$ for $0<x<1$

Question

I want to show that the infinite product $$\displaystyle\prod _{n=1} ^\infty (1+x^n)$$ converges for $0<x<1$. This is easy when we use the theory of infinite products in Complex Analysis. Indeed, we have $$\displaystyle\prod _{n=1} ^\infty (1+x^n) ~\textrm{converges (absolutely)} \Leftrightarrow \sum _{n=1}^\infty x^n ~\text{converges (absolutely)}$$ and the latter clearly holds. But I'm wondering that, is there a more elementary or direct proof of the question? Thanks in advance.

Answer 1

The relation between (absolute) convergence of $\prod_{n=1} ^\infty (1+a_n)$ and $\sum _{n=1}^\infty a^n$ is based on the fact that $\log(1+t) \sim t$, or $e^t \sim 1+t$, for small $t$.

For $\prod _{n=1} ^\infty (1+x^n)$ with $0 < x < 1$ one can avoid the estimation of the error terms in those approximations and use that $1+t \le e^t$ for real $t$. It follows that $$ \prod_{k=1}^n (1+x^k) \le \prod_{k=1}^n e^{x^k} = e^{\sum_{k=1}^n x^k} \le e^{1/(1-x)} \,, $$ i.e. the sequence of partial products is increasing and bounded above, and therefore convergent.

Answer 2

It folloows by taking logarithm and using the fact $\frac {\log (1+t)} t \to 1$ as $ t \to 0+$.