How do I prove that $\prod\limits_{n=1}^{\infty} (1-x^n)$ is absolutely convergent for $0 \leq x < 1$?

Question

Consider the infinite product

$$\prod_{n=1}^{\infty} (1-x^n).$$

Prove that the above infinite product is absolutely convergent for $0 \leq x < 1.$

I have considered the sequence of partial products $\{F_m(x) \}$ where $$F_m (x) = \prod_{n=1}^{m} (1-x^n).$$ Then I have observed that $F_{m+1} (x) = (1 - x^{m+1}) F_m (x).$ If $0 \leq x <1$ then $F_{m+1} (x) \leq F_m (x).$ So the sequence of partial products is monotone decreasing and bounded below by $0.$ So for every $0 \leq x < 1$ the sequence $\{F_m(x) \}$ is monotone decreasing and bounded below by $0$ and therefore it is convergent. Since for $0 \leq x < 1$ the sequence $\{F_m (x) \}$ is a sequence of positive terms (as each term is the product of finitely many positive terms) so we can conclude that the sequence $\{F_m (x) \}$ is absolutely convergent for each $0 \leq x < 1.$ Therefore for each $0 \leq x <1$ the infinite product $\prod_{n=1}^{\infty} (1-x^n)$ is absolutely convergent as well.

Is the above argument ok? Please verify it.

Thank you very much for your valuable time.

Answer 1

According to the definition, the infinite product $\prod\limits_{n=1}^{\infty} (1+a_n)$ is absolutely convergent if $\prod\limits_{n=1}^{\infty} (1 + |a_n|)$ is convergent. Now observe that $\prod\limits_{n=1}^{\infty} (1 + |a_n|)$ is convergent iff $\sum\limits_{n=1}^{\infty} \log (1 + |a_n|)$ is convergent. Also observe that $\log (1 + |a_n|) \leq |a_n|,$ for all $n \in \Bbb N.$ Therefore we can conclude that the infinite product $\prod\limits_{n=1}^{\infty} (1 + a_n)$ is absolutely convergent iff the infinite series $\sum\limits_{n=1}^{\infty} a_n$ is absolutely convergent.

In this case $a_n = - x^n,$ for all $n \in \Bbb N.$ So in order to show that the given infinite product $\prod\limits_{n=1}^{\infty} (1-x^n)$ is absolutely convergent for each $0 \leq x < 1$ we need only to show that the infinite series $\sum\limits_{n=1}^{\infty} x^n$ is convergent for each $0 \leq x <1.$ Which is a very well known result and I leave the details for you to verify.

Answer 2

May be this can help:

$$\prod_{n=1}^{m}{(1-x^n)}=\exp\left(\sum_{n=1}^{m}{\ln(1-x^n)}\right)$$ $$= \exp\left(-\sum_{n=1}^{m}{\sum_{\sigma=1}^{\infty}{\frac{x^{\sigma n}}{\sigma}}}\right)=\exp\left(-\sum_{\sigma=1}^{\infty}{\frac{1}{\sigma}}\sum_{n=1}^{m}{x^{\sigma n}}\right) $$