TITLE: Product of all numbers in a given interval $[n,m]$
QUESTION [14 upvotes]: Introduction
I was wondering if, you define some interval $[n,m]$ which contains all numbers between $n$ and $m$ (consider either $\mathbb Q$ or $\mathbb R$), what would be the product of all those numbers? 
One of my instant guesses was that'll tend to infinity.
Clearly, it would be neat to assume $a\in[n,m], a\ne 0$ since the product will be $0$ otherwise.
In addition, $n,m\ge0$ since multiplying infinitely many negative numbers is undefined.


Q: Is it correct to claim that interval type $[a\gt0,b\gt a]$ in its
  product of all of its infinitely many real numbers, tends to $0$ no
  matter how big $b$ is, since you can include numbers infinitely close to $0$?
A: It seems it can either tend to $0$, $\infty$ or a finite value $x$, which is analyzed below. 


Interval of type $[a\gt1,b\gt a]$, would tend to infinity in all cases since we never decrease in value no matter what number we pick to multiply next.

Symmetrical expression to work multiply the numbers
Imagine taking a interval $[n,m]$, and taking the numbers for each $k$ iteration, as averages of previous numbers. Like this:
$$n,m$$
$$n,\frac{n+m}{2},m$$
$$n,\frac{3n+m}{4},\frac{n+m}{2},\frac{n+3m}{4},m$$
$$\dots$$
Which would for $k$ iterations as $k$ tends towards $\infty$, eventually contain all numbers in that interval.
The product of all those numbers for $k$ iterations can be written as:
$$ \prod ^{2^k} _{x=0} \frac{(2^k - x)n+xm}{2^k} $$
If we now take an interval $[n,m]$ as $[2,3]$ for example, we have:
For $k=4$ it is: $5.11501\dots \times 10^6$
 For $k=8$ it is: $3.24661\dots \times 10^{101}$
 For $k=10$ it is: $7.56107\dots \times 10^{404}$
 And we can say it clearly tends towards $\infty$ itself.
And if we pick an interval containing numbers less than 1, we can see it can tend towards $0$;
For example: $[0.1,2],k=14$ then: $1.3926\dots \times 10^{-1062}$
But If we have a greater number, we can still tend towards infinity;
For example: $[0.1,3],k=14$ then: $8.6083\dots \times 10^{1535}$

But then If we choose $[0.1,2.3639556]$ then for $k=14$ we will get:
  $1.000756\dots$  Seems it tends to a bit over $1$?
Q: Does that mean we can by choosing the right ratio, or the right
  values in the interval, make it tend towards any value?
A: But then If we increase the $k$, we start tending to either $\infty$ or $0$ very rapidly; for $k=16$ I had to decrease the $m$ a
  bit to $2.3638689$ for it to be roughly $1.0076\dots$

This brings up the question;

Can we actually make the product tend to a finite value for the case
  where $k$ approaches $\infty$ by choosing the right ratio of $n$ and
  $m$ in which $(n \lt 1)$ , $(m\gt 1)$?


Calculating values for finite cases?
I've now tried to find the $m$ for $[0.1,m]$ so that the product of the interval ($x$) :
For $k=2$, $m\approx 3.10745$ for it to be $x\approx 1.00001$
For $k=10$, $m= 2.36569706$ for it to be $x= 1$
For $k=14$, $m\approx 2.363955478620$ for it to be $x\approx 1.00000001474698$
For $k=15$, $m\approx 2.363897553300$ for it to be $x\approx 1.00000169779107$
For $k=16$, $m\approx 2.363868593513$ for it to be $x\approx 1.00000000497575$
$$\dots$$
As you can see, for bigger $k$ we get more precision on the digits of $m$
This brings up another question;

How would one compute $m$ with most precision for choosen $n$,
  $(0<n<1)$ where $k$ tends to $\infty$ so that the product of all reals
  in $[n,m]$ is a finite value $x\in \mathbb R$?

Do you have any software, any code for a computing algorithm, or a better formula than my symmetrical expression in mind? To preform the calculation where $k$ tends to $\infty$?

Summing Everything up
When calculating the infinite product of all real numbers in the interval $[n,m]$, $(n\lt m)$, We have a few cases we can look at individually:
$1.$ When $n\lt0$ the result is "undefined" since we have $\infty$ many negative numbers of which we can extract $(-1)$, which can then be written as $\lim\limits_{n\to\infty}(-1)^n$ which is undefined.
$2.$ When $a\in[n,m], a=0$ The result is clearly $0$
$3.$ When $0<n<m<1$ then the result tends to $0$
$4.$ When $n>1,m>n$ the result tends to $\infty$
$5.$ When $0<n<1$ and $m>1$ The result can either tend to $0$, $\infty$ or a finite value $x$, which can be calculated by using a symmetric expression I used in my approach.

Better Calculations

I use WolframAlpha while increasing $k$ but it runs up to $20$ since bigger $k$ exceed its computational time allowed.

When applying calculations proposed by String in his answer (using Newton-Raphson method), and feeding them to WolframAlpha, the most precision we get is (for $x=1$):
$n=\frac{1}{2}$, $m\approx 1.6030164899169670747912206652529016572070546450201637$
$n=\frac{1}{3}$, $m\approx 1.8699324270643973162008471123760292568887231725646945$
$n=\frac{1}{4}$, $m\approx 2.0245006935913233776633741232407065465057011574013821$
$n=\frac{1}{5}$, $m\approx 2.1267864647345386651244066521323021196320323179442441$
$$\dots$$
The $m$ seems to tend to $e$ as $n\to\infty$.

I might get back to this if I get a chance, but I doubt it'll be soon enough.

REPLY [4 votes]: There is no such thing as a product of uncountably many numbers (where not most of them are $=1$ or some are $=0$). Compare to sums: Even with countably many summnds, we do not speak of sums, but of series (even though we suggestively use the same symbol $\sum$ for both). Those have very different properties from sums: A sum of rationals is always defined, is always a rational, and does not depend on the order of summation. On the other hand, a series of rationals may fail to converge, or converge to an irrational number, or converge to different values if we change the order of the terms. So to repeat: Even a "sum" of countably many numbers is not really a sum. A "sum" of uncountably many (non-zero) summands  is even more horrible: For any such best there must exist some $\epsilon>0$ such that uncountably many terms are $>\epsilon$, or uncountzably many terms are $<-\epsilon$; already their contribution is (positive or negative) infinite. It is not an easy task, especially for an arbitrary index set, to assign any meaning to this. The same argument holds for products (if not with some extra considerations).
That being said, you gave a specific definition of an expression $$P(a,b)=\lim_{n\to\infty}P(a,b;2^n),$$
where
$$P(a,b;N)=\prod_{j=0}^N\frac{(N-j)a+jb}{N}, $$ which we shall investigate (and better forget that we want to call this "product of all numbers in $[a,b]$).
Consider first the case $0<a<b$.
$$\frac1N\ln P(a,b;N) = \sum_{j=0}^N\frac1N\ln \frac{(N-j)a+jb}{N},$$
which looks a lot like a Riemann sum. Indeed, as $\ln$ is strictly increasing, we see that
$$ \frac1N\ln P(a,b;N)=\int_a^b\ln x\,\mathrm dx+R_n$$
where $\frac 1N\ln a<R_n<\frac 1N\ln b$. As $x(\ln x-1)$ is an anti-derivative fo the logarithm, we arrive at
$$ P(a,b;N) = \bigl(e^{b(\ln b-1)-a(\ln a-1)}\bigr)^N\cdot c_N$$
with $a<c_N<b$. The limit as $N\to\infty$ can only exist if $$b(\ln b-1)=a(\ln a-1).$$
Such pairs $(a,b)$ are rare: The function $x\mapsto x(\ln x-1)$ decreases from $1$ to $0$ on $(0,1]$, and then increases, reaching $1$ at $x=e$. Hence the only suitable pairs $(a,b)$ are with $0<a<1$ and one suitable $1<b<e$.
By investigating $c_N$ closer (comparing with the trapezoidal rule), one arrives at
$P(a,b)=\sqrt ab$ for such $a,b$.
Interestingly, if $P(a,b)$ exists, we also have $P(-b,-a)=-P(a,b)$. This is because the specific definition of $P(a,b)$ uses only $P(a,b; 2^n)$ and for $n>0$ we have $P(a,b,2^n)=P(-b,-a,2^n)$. (Though for me this fun result is more a hint that the definition of $P$ is not perfect)