0
$\begingroup$

Can the infinite product $\prod_{n=1}^\infty \frac{a_n}{a_n + 1}$ converge or diverge to zero when the sequence $a_n$ is monotonically increasing? I conjecture that this is impossible given that $\lim_{n\to\infty}\frac{a_n}{a_n+1}$ by the monotonic increase of the $a_n$. However, I don't know of a rigorous proof or disproof of this, and analyzing convergence of infinite series and infinite products is not something I am very familiar with. If someone could provide a proof or disproof of this conjecture (preferably with a reference to a known theorem), or explain what additional restrictions need to be imposed on $a_n$ for the conjecture to hold, I would be very appreciative!

$\endgroup$
1
  • 1
    $\begingroup$ Problems about the "divergence to zero" of an infinite product are often successfully attacked by taking the logarithm to obtain an infinite sum (series). See this earlier Question on that topic. $\endgroup$ – hardmath Nov 17 '20 at 4:19
1
$\begingroup$

Let $q_n=\dfrac{1}{1+a_n}$ be such that $0\lt q_n\lt 1,$ then by the elementary theory of infinite products $\prod_{n=1}^\infty (1-q_n)$ converges to a non-zero number if and only if $\sum_{n=1}^{\infty}q_n$ converges.
This clearly gives you a criteria to find the convergence to a non-zero number.

For example: if you set $a_n$ to be the $n$th prime (as mentioned in your comment), then $\sum_{n=1}^{\infty}q_n$ diverges and therefore ....

$\endgroup$
4
  • $\begingroup$ There are a few things that I am confused about here. First, your source seems to indicate that we would need $\prod_{n=1}^\infty log \frac{a_n}{a_n + 1}$ rather than $\prod_{n=1}^\infty (1-q_n)$. Also, Diego Zhang and @achille hui indicate that the product converges, while your criterion indicates it diverges. Can anyone help explain this? $\endgroup$ – Math Rules Nov 17 '20 at 5:52
  • $\begingroup$ @MathRules: If you read the Wikipedia page carefully, you will see the criteria for convergence of $\prod_{n=1}^\infty (1-q_n).$ Also what achille hui proved is the product "diverge to zero." Therefore does not contradict with what I have wrote in my answer. $\endgroup$ – Bumblebee Nov 17 '20 at 5:57
  • $\begingroup$ You are correct. I thought you were referring to formulas at the top of the page, my mistake! The criteria for $\prod_{n=1}^\infty 1-q_n$ is indeed there, I just had missed it. $\endgroup$ – Math Rules Nov 17 '20 at 6:01
  • $\begingroup$ @MathRules: I'm glad, if I was able to help. $\endgroup$ – Bumblebee Nov 17 '20 at 6:05
3
$\begingroup$

Every term is less than $1$, so it converges to a constant less than 1.

Take $a_n=n$, and the product converge to $0$.

Take $a_n=2^n$, and the product converges, but not to $0$. This can be obtained by taking the logarithm and using the inequality $\ln(1+x)<x$.

$\endgroup$
2
  • $\begingroup$ What happens if $a_n$ is the $n$th prime? Could we take the logarithm and then apply the Prime Number Theorem (or some other theorem) to show the product converges to a nonzero value? $\endgroup$ – Math Rules Nov 17 '20 at 5:00
  • 1
    $\begingroup$ @MathRules If $a_n$ is $n^{th}$ prime, the limit is $0$. Loosely speaking, $\prod_{n=1}^\infty \frac{p_n}{p_n+1} = \prod_{n=1}^\infty \frac{1}{1+ p_n^{-1}} = \prod_{n=1}^\infty\frac{1 - p_n^{-1}}{1 - p_n^{-2}} = \frac{\zeta(2)}{\zeta(1)} = 0$ $\endgroup$ – achille hui Nov 17 '20 at 5:47

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.