Is the following infinite product: $$ \prod_{\substack{(a,b) \in \mathbb{Z}^2 \\ a > b}} \frac{x+b}{x+a} $$ defined? If so, does it simplify to an exponential function of $x$?

The subscript is over all integer pairs $a$ and $b$ such that $a>b $

Motivation/suggestions: When plotting a graph of a finite truncation of the product :enter image description here it resembles a negative exponential function. However such an approximation appears to not be accurate for small values of $x$ (where it has vertical asymptotes).

Can it be transformed in such a way that it becomes an exponential function, by modifying each term individually so that the whole function is finite and the asymptotes lie at infinite distance in the limit?

Or can we show that the derivative of this infinite product with respect to $x$ is a constant factor of itself or not, thus proving that it is an exponential?

I believe it should be exponential because it was obtained by raising $e$ to the power a modified version of an integral of a constant function.

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    $\begingroup$ How to ask a good question. After reading that, you are welcome to return to this post to improve it. Don't expect an answer, unless you improve the question, and don't expect an answer in a day or two. The better the question, the more likely it will be answered. $\endgroup$
    – amWhy
    Commented Feb 9, 2022 at 17:23
  • $\begingroup$ Can you explain what's wrong with it, other than the lack of MathJax ; which other criteria it doesn't meet, and how I can expect an answer at all if it is already buried beneath thousands of more recent questions? My other questions have received answers within 1 hour of being posted, but presumably that is precisely why they received any at all. $\endgroup$ Commented Feb 9, 2022 at 18:27
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    $\begingroup$ A good edit, one good edit, will renew the question in the list of questions. Please do improve/explain like you hope to. It will help resurface the question, so it's not buried in beneath much at all. And you asked this only yesterday. Improving it today, if you follow suggestions in the link I provided, will give it a fairly good chance of being reviewed. You can ping me, if you'd like, to check on your improvement. We, or at least I, sincerely seek to help users be successful on this site. $\endgroup$
    – amWhy
    Commented Feb 9, 2022 at 18:33
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    $\begingroup$ Also, @Amoeba, sometimes there is no rhyme or reason for why some questions get answered quickly, and others don't. Everyone's schedules vary. Please never take a day or two since a question is posted, as any indicator or anything. Sometimes it is only because people here are busy, often outside of math.se. Also keep in mind that everyone, including answerers, and moderators, are strictly volunteers on SE sites, including this site. So we have jobs, some attend school, etc.. Cheers! $\endgroup$
    – amWhy
    Commented Feb 9, 2022 at 19:42
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    $\begingroup$ You assume way too much, starting with the idea that your infinite product exists. Also, it obviously has many terms that can be canceled. What if we cancel them? What is it that remains? $\endgroup$ Commented Feb 12, 2022 at 8:50

1 Answer 1


This infinite product definitely doesn't converge for any $x$. A necessary condition for an infinite product over $a,b$ to converge is that the individual factors tend to $1$ as $|a|,|b|\to\infty$, but that's not the case here (since $|a|$ and $|b|$ can tend to $\infty$ together in lots of different ways, for example with $b>0$ and $a=2b$).

If the intended expression is a limit of a specific sequence of finite products, that might be a different story due to cancellation of some sort; it would depend on the specific sequence.

  • $\begingroup$ "A necessary condition for an infinite product over a,b to converge is that the individual factors tend to 1 as |a|,|b|→∞, but that's not the case here " Yes I know this, that is why I wrote : 'Is there a way to transform the infinite product so that it becomes an exponential function, e.g. by modifying each term individually?' I was asking if each factor or maybe the whole series can be divided by a constant factor each time the truncation is lengthened so that it doesn't diverge. $\endgroup$ Commented Feb 13, 2022 at 12:27
  • $\begingroup$ You might northave got this impression because Martin Sleziak edited it... $\endgroup$ Commented Feb 13, 2022 at 12:33
  • $\begingroup$ Can you at least explain why the question was closed? $\endgroup$ Commented Feb 13, 2022 at 12:48

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