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

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.

• Do you know the definition of an absolute convergence of a product? – Wojowu Mar 22 at 16:59
• If you multiply two numbers, both in the interval $(0,1]$, in what interval does the end-result sit? Do you know what induction is? – samerivertwice Mar 22 at 17:00
• I don't know if this would help but you can use pentagonal number theorem to simplify the product – Rohan Shinde Mar 22 at 17:02
• @Darkai I haven't studied Euler's pentagonal number theorem. But yes I am now studying the subject partition theory. – math maniac. Mar 22 at 17:04
• @Wojowu I don't know that. Perhaps that was the problem when I did the solution. That is why I post my solution here as I was bit unsure about my solution. Can you tell me what is meant by absolute convergence of products? Is it different from what we know about the sums? – math maniac. Mar 22 at 17:06

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.
$$\prod_{n=1}^{m}{(1-x^n)}=\exp\left(\sum_{n=1}^{m}{\ln(1-x^n)}\right)$$ $$= \exp\left(-\sum_{n=1}^{m}{\sum_{\sigma=1}^{\infty}{\frac{x^{\sigma n}}{\sigma}}}\right)=\exp\left(-\sum_{\sigma=1}^{\infty}{\frac{1}{\sigma}}\sum_{n=1}^{m}{x^{\sigma n}}\right)$$
• Now \begin{align} \exp \left (- \sum\limits_{\sigma = 1}^{\infty} \frac {1} {\sigma} \sum\limits_{n=1}^{m} x^{\sigma n} \right ) & = \exp \left ( -\sum\limits_{\sigma = 1}^{\infty} \frac {1} {\sigma} \cdot \frac {x^{\sigma} (1 - x^{\sigma m})} {1 - x^{\sigma}} \right ).\end{align} What can be done then @HAMIDINE SOUMARE? – math maniac. Mar 22 at 18:12