3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Do you know the definition of an absolute convergence of a product? $\endgroup$ – Wojowu Mar 22 at 16:59
  • $\begingroup$ 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? $\endgroup$ – samerivertwice Mar 22 at 17:00
  • $\begingroup$ I don't know if this would help but you can use pentagonal number theorem to simplify the product $\endgroup$ – Rohan Shinde Mar 22 at 17:02
  • $\begingroup$ @Darkai I haven't studied Euler's pentagonal number theorem. But yes I am now studying the subject partition theory. $\endgroup$ – math maniac. Mar 22 at 17:04
  • $\begingroup$ @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? $\endgroup$ – math maniac. Mar 22 at 17:06
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

May be this can help:

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – math maniac. Mar 22 at 18:12
  • $\begingroup$ Can you say what was wrong in my steps @HAMIDINE SOUMARE in the body of the question? $\endgroup$ – math maniac. Mar 22 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.